Content check — Algebra and Trig, Chapter 6: Exponential and Logarithmic Functions

Try It #1 §6.1 source ↗

Which of the following equations represent exponential functions?

• $f (x) =$ $2 x^{2} -$ $3 x +$ $1$

• $g (x) = {0.875}^{x}$

• $h (x) =$ $1.75 x +$ $2$

• $j (x) =$ ${1095.6}^{- 2 x}$

Answer

g (x) = {0.875}^{x}

and

j (x) =

{1095.6}^{- 2 x}

represent exponential functions.

Try It #2 §6.1 source ↗

Let

f (x) =

8 {(1.2)}^{x - 5} .

Evaluate

f (3)

using a calculator. Round to four decimal places.

Answer

5.5556

Try It #3 §6.1 source ↗

The population of China was about 1.39 billion in the year 2013, with an annual growth rate of about

0.6 % .

This situation is represented by the growth function

P (t) =

1.39 {(1.006)}^{t},

where

t

is the number of years since

2013.

To the nearest thousandth, what will the population of China be for the year 2031? How does this compare to the population prediction we made for India in Example 3?

Answer

About

1.548

billion people; by the year 2031, India’s population will exceed China’s by about 0.001 billion, or 1 million people.

Try It #4 §6.1 source ↗

A wolf population is growing exponentially. In 2011,

129

wolves were counted. By

2013,

the population had reached 236 wolves. What two points can be used to derive an exponential equation modeling this situation? Write the equation representing the population

N

of wolves over time

t .

Answer

(0, 129)

and

(2, 236);

N (t) =

129 {(1 .3526)}^{t}

Try It #5 §6.1 source ↗

Given the two points

(1, 3)

and

(2, 4.5),

find the equation of the exponential function that passes through these two points.

Answer

f (x) =

2 {(1.5)}^{x}

Try It #6 §6.1 source ↗

Find an equation for the exponential function graphed in Figure 6.

Graph of an increasing function with a labeled point at (0, sqrt(2)).

Figure 6

Answer

f (x) = \sqrt{2} {(\sqrt{2})}^{x} .

Answers may vary due to round-off error. The answer should be very close to

1.4142 {(1.4142)}^{x} .

Try It #7 §6.1 source ↗

Use a graphing calculator to find the exponential equation that includes the points (3, 75.98) and (6, 481.07).

Answer

y \approx 12 \cdot {1.85}^{x}

Try It #8 §6.1 source ↗

An initial investment of $100,000 at 12% interest is compounded weekly (use 52 weeks in a year). What will the investment be worth in 30 years?

Answer

about $3,644,675.88

Try It #9 §6.1 source ↗

Refer to Example 9. To the nearest dollar, how much would Lily need to invest if the account is compounded quarterly?

Answer

$13,693

Try It #10 §6.1 source ↗

Use a calculator to find

e^{- 0.5} .

Round to five decimal places.

Answer

e^{- 0.5} \approx

0.60653

Try It #11 §6.1 source ↗

A person invests $100,000 at a nominal 12% interest per year compounded continuously. What will be the value of the investment in 30 years?

Answer

$3,659,823.44

Try It #12 §6.1 source ↗

Using the data in Example 12, how much radon-222 will remain after one year?

Answer

3.77E-26 (This is calculator notation for the number written as

3.77 \times 10^{- 26}

in scientific notation. While the output of an exponential function is never zero, this number is so close to zero that for all practical purposes we can accept zero as the answer.)

6.1-1 §6.1 source ↗

Explain why the values of an increasing exponential function will eventually overtake the values of an increasing linear function.

Answer

Linear functions have a constant rate of change. Exponential functions increase based on a percent of the original.

6.1-3 §6.1 source ↗

The Oxford Dictionary defines the word nominal as a value that is “stated or expressed but not necessarily corresponding exactly to the real value.”³ Develop a reasonable argument for why the term nominal rate is used to describe the annual percentage rate of an investment account that compounds interest.

Answer

When interest is compounded, the percentage of interest earned to principal ends up being greater than the annual percentage rate for the investment account. Thus, the annual percentage rate does not necessarily correspond to the real interest earned, which is the very definition of nominal.

6.1-5 §6.1 source ↗

For the following exercises, identify whether the statement represents an exponential function. Explain.

A population of bacteria decreases by a factor of $\frac{1}{8}$ every $24$ hours.

Answer

exponential; the population decreases by a proportional rate. .

6.1-7 §6.1 source ↗

For the following exercises, identify whether the statement represents an exponential function. Explain.

For each training session, a personal trainer charges his clients $$ 5$ less than the previous training session.

Answer

not exponential; the charge decreases by a constant amount each visit, so the statement represents a linear function. .

6.1-9 §6.1 source ↗

For the following exercises, consider this scenario: For each year $t,$ the population of a forest of trees is represented by the function $A (t) =$ $115 {(1.025)}^{t} .$ In a neighboring forest, the population of the same type of tree is represented by the function $B (t) =$ $82 {(1.029)}^{t} .$ (Round answers to the nearest whole number.)

Which forest’s population is growing at a faster rate?

Answer

The forest represented by the function

B (t) =

82 {(1.029)}^{t} .

6.1-11 §6.1 source ↗

For the following exercises, consider this scenario: For each year $t,$ the population of a forest of trees is represented by the function $A (t) =$ $115 {(1.025)}^{t} .$ In a neighboring forest, the population of the same type of tree is represented by the function $B (t) =$ $82 {(1.029)}^{t} .$ (Round answers to the nearest whole number.)

Assuming the population growth models continue to represent the growth of the forests, which forest will have a greater number of trees after $20$ years? By how many?

Answer

After

t = 20

years, forest A will have

43

more trees than forest B.

6.1-13 §6.1 source ↗

For the following exercises, consider this scenario: For each year $t,$ the population of a forest of trees is represented by the function $A (t) =$ $115 {(1.025)}^{t} .$ In a neighboring forest, the population of the same type of tree is represented by the function $B (t) =$ $82 {(1.029)}^{t} .$ (Round answers to the nearest whole number.)

Discuss the above results from the previous four exercises. Assuming the population growth models continue to represent the growth of the forests, which forest will have the greater number of trees in the long run? Why? What are some factors that might influence the long-term validity of the exponential growth model?

Answer

Answers will vary. Sample response: For a number of years, the population of forest A will increasingly exceed forest B, but because forest B actually grows at a faster rate, the population will eventually become larger than forest A and will remain that way as long as the population growth models hold. Some factors that might influence the long-term validity of the exponential growth model are drought, an epidemic that culls the population, and other environmental and biological factors.

6.1-15 §6.1 source ↗

For the following exercises, determine whether the equation represents exponential growth, exponential decay, or neither. Explain.

$y =$ $220 {(1.06)}^{x}$

Answer

exponential growth; The growth factor,

1.06,

is greater than

1.

6.1-17 §6.1 source ↗

For the following exercises, determine whether the equation represents exponential growth, exponential decay, or neither. Explain.

$y =$ $11,$ $701 {(0.97)}^{t}$

Answer

exponential decay; The decay factor,

0.97,

is between

0

and

1.

6.1-19 §6.1 source ↗

For the following exercises, find the formula for an exponential function that passes through the two points given.

$(0, 2000)$ and $(2, 20)$

Answer

f (x) =

2000 {(0.1)}^{x}

6.1-21 §6.1 source ↗

For the following exercises, find the formula for an exponential function that passes through the two points given.

$(- 2, 6)$ and $(3, 1)$

Answer

f (x) =

{(\frac{1}{6})}^{- \frac{3}{5}} {(\frac{1}{6})}^{\frac{x}{5}} \approx

2.93 {(0.699)}^{x}

6.1-23 §6.1 source ↗

For the following exercises, determine whether the table could represent a function that is linear, exponential, or neither. If it appears to be exponential, find a function that passes through the points.

$x$	1	2	3	4
$f (x)$	70	40	10	-20

Answer

Linear

6.1-25 §6.1 source ↗

For the following exercises, determine whether the table could represent a function that is linear, exponential, or neither. If it appears to be exponential, find a function that passes through the points.

$x$	1	2	3	4
$m (x)$	80	61	42.9	25.61

Answer

Neither

6.1-27 §6.1 source ↗

For the following exercises, determine whether the table could represent a function that is linear, exponential, or neither. If it appears to be exponential, find a function that passes through the points.

$x$	1	2	3	4
$g (x)$	-3.25	2	7.25	12.5

Answer

Linear

6.1-29 §6.1 source ↗

For the following exercises, use the compound interest formula, $A (t) =$ $P {(1 + \frac{r}{n})}^{n t} .$

What was the initial deposit made to the account in the previous exercise?

Answer

$ 10, 250

6.1-31 §6.1 source ↗

For the following exercises, use the compound interest formula, $A (t) =$ $P {(1 + \frac{r}{n})}^{n t} .$

An account is opened with an initial deposit of $6,500 and earns $3.6 %$ interest compounded semi-annually. What will the account be worth in $20$ years?

Answer

$ 13, 268.58

6.1-33 §6.1 source ↗

For the following exercises, use the compound interest formula, $A (t) =$ $P {(1 + \frac{r}{n})}^{n t} .$

Solve the compound interest formula for the principal, $P$ .

Answer

P =

A (t) \cdot {(1 + \frac{r}{n})}^{- n t}

6.1-35 §6.1 source ↗

For the following exercises, use the compound interest formula, $A (t) =$ $P {(1 + \frac{r}{n})}^{n t} .$

How much more would the account in the previous two exercises be worth if it were earning interest for $5$ more years?

Answer

$ 4,572.56

6.1-37 §6.1 source ↗

For the following exercises, use the compound interest formula, $A (t) =$ $P {(1 + \frac{r}{n})}^{n t} .$

Use the formula found in the previous exercise to calculate the interest rate for an account that was compounded semi-annually, had an initial deposit of $9,000 and was worth $13,373.53 after 10 years.

Answer

4 %

6.1-39 §6.1 source ↗

For the following exercises, determine whether the equation represents continuous growth, continuous decay, or neither. Explain.

$y =$ $3742 {(e)}^{0.75 t}$

Answer

continuous growth; the growth rate is greater than

0.

6.1-41 §6.1 source ↗

For the following exercises, determine whether the equation represents continuous growth, continuous decay, or neither. Explain.

$y =$ $2.25 {(e)}^{- 2 t}$

Answer

continuous decay; the growth rate is less than

0.

6.1-43 §6.1 source ↗

For the following exercises, determine whether the equation represents continuous growth, continuous decay, or neither. Explain.

How much less would the account from Exercise 42 be worth after $30$ years if it were compounded monthly instead?

Answer

$ 669.42

6.1-45 §6.1 source ↗

For the following exercises, evaluate each function. Round answers to four decimal places, if necessary.

$f (x) =$ $-$ $4^{2 x + 3},$ for $f (- 1)$

Answer

f (- 1) = - 4

6.1-47 §6.1 source ↗

For the following exercises, evaluate each function. Round answers to four decimal places, if necessary.

$f (x) =$ $-$ $2 e^{x - 1},$ for $f (- 1)$

Answer

f (- 1) \approx

-

0.2707

6.1-49 §6.1 source ↗

For the following exercises, evaluate each function. Round answers to four decimal places, if necessary.

$f (x) =$ $1.2 e^{2 x} -$ $0.3,$ for $f (3)$

Answer

f (3) \approx

483.8146

6.1-51 §6.1 source ↗

For the following exercises, use a graphing calculator to find the equation of an exponential function given the points on the curve.

$(0, 3)$ and $(3, 375)$

Answer

y = 3 \cdot 5^{x}

6.1-53 §6.1 source ↗

For the following exercises, use a graphing calculator to find the equation of an exponential function given the points on the curve.

$(20, 29.495)$ and $(150, 730.89)$

Answer

y \approx 18 \cdot {1.025}^{x}

6.1-55 §6.1 source ↗

For the following exercises, use a graphing calculator to find the equation of an exponential function given the points on the curve.

$(11,310.035)$ and $(25,356365.2)$

Answer

y \approx 0.2 \cdot {1.95}^{x}

6.1-57 §6.1 source ↗

For the following exercises, use a graphing calculator to find the equation of an exponential function given the points on the curve.

Repeat the previous exercise to find the formula for the APY of an account that compounds daily. Use the results from this and the previous exercise to develop a function $I (n)$ for the APY of any account that compounds $n$ times per year.

Answer

APY =

\frac{A (t) - a}{a} =

\frac{a {(1 + \frac{r}{365})}^{365 (1)} - a}{a} =

\frac{a [{(1 + \frac{r}{365})}^{365} - 1]}{a} =

{(1 + \frac{r}{365})}^{365} -

1;

I (n) =

{(1 + \frac{r}{n})}^{n} -

1

6.1-59 §6.1 source ↗

For the following exercises, use a graphing calculator to find the equation of an exponential function given the points on the curve.

In an exponential decay function, the base of the exponent is a value between 0 and 1. Thus, for some number $b > 1,$ the exponential decay function can be written as $f (x) =$ $a \cdot {(\frac{1}{b})}^{x} .$ Use this formula, along with the fact that $b = e^{n},$ to show that an exponential decay function takes the form $f (x) =$ $a {(e)}^{- n x}$ for some positive number $n$ .

Answer

Let

f

be the exponential decay function

f (x) =

a \cdot {(\frac{1}{b})}^{x}

such that

b > 1.

Then for some number

n > 0,

f (x) =

a \cdot {(\frac{1}{b})}^{x} =

a {(b^{- 1})}^{x} =

a {({(e^{n})}^{- 1})}^{x} =

a {(e^{- n})}^{x} =

a {(e)}^{- n x} .

6.1-61 §6.1 source ↗

For the following exercises, use a graphing calculator to find the equation of an exponential function given the points on the curve.

The fox population in a certain region has an annual growth rate of 9% per year. In the year 2012, there were 23,900 fox counted in the area. What is the fox population predicted to be in the year 2020?

Answer

47, 622

fox

6.1-63 §6.1 source ↗

For the following exercises, use a graphing calculator to find the equation of an exponential function given the points on the curve.

In the year 1985, a house was valued at $110,000. By the year 2005, the value had appreciated to $145,000. What was the annual growth rate between 1985 and 2005? Assume that the value continued to grow by the same percentage. What was the value of the house in the year 2010?

Answer

1.39 %;

$ 155, 368.09

6.1-65 §6.1 source ↗

For the following exercises, use a graphing calculator to find the equation of an exponential function given the points on the curve.

Jaylen wants to save $54,000 for a down payment on a home. How much will he need to invest in an account with 8.2% APR, compounding daily, in order to reach his goal in 5 years?

Answer

$ 35, 838.76

6.1-67 §6.1 source ↗

For the following exercises, use a graphing calculator to find the equation of an exponential function given the points on the curve.

Alyssa opened a retirement account with 7.25% APR in the year 2000. Her initial deposit was $13,500. How much will the account be worth in 2025 if interest compounds monthly? How much more would she make if interest compounded continuously?

Answer

$ 82, 247.78;

$ 449.75

Try It #1 §6.2 source ↗

Sketch the graph of

f (x) = 4^{x} .

State the domain, range, and asymptote.

Answer

The domain is $(- \infty, \infty);$ the range is $(0, \infty);$ the horizontal asymptote is $y = 0.$

Try It #2 §6.2 source ↗

Graph

f (x) =

2^{x - 1} +

3.

State domain, range, and asymptote.

Answer

The domain is $(- \infty, \infty);$ the range is $(3, \infty);$ the horizontal asymptote is $y = 3.$

Graph of the function, f(x) = 2^(x-1)+3, with an asymptote at y=3. Labeled points in the graph are (-1, 3.25), (0, 3.5), and (1, 4).

Try It #3 §6.2 source ↗

Solve

4 =

7.85 {(1.15)}^{x} -

2.27

graphically. Round to the nearest thousandth.

Answer

x \approx - 1.608

Try It #4 §6.2 source ↗

Sketch the graph of

f (x) =

\frac{1}{2} {(4)}^{x} .

State the domain, range, and asymptote.

Answer

The domain is $(- \infty, \infty);$ the range is $(0, \infty);$ the horizontal asymptote is $y = 0.$

Graph of the function, f(x) = (1/2)(4)^(x), with an asymptote at y=0. Labeled points in the graph are (-1, 0.125), (0, 0.5), and (1, 2).

Try It #5 §6.2 source ↗

Find and graph the equation for a function,

g (x),

that reflects

f (x) = {1.25}^{x}

about the y-axis. State its domain, range, and asymptote.

Answer

The domain is $(- \infty, \infty);$ the range is $(0, \infty);$ the horizontal asymptote is $y = 0.$

Graph of the function, g(x) = -(1.25)^(-x), with an asymptote at y=0. Labeled points in the graph are (-1, 1.25), (0, 1), and (1, 0.8).

Try It #6 §6.2 source ↗

Write the equation for function described below. Give the horizontal asymptote, the domain, and the range.

• $f (x) = e^{x}$ is compressed vertically by a factor of $\frac{1}{3},$ reflected across the x-axis and then shifted down $2$ units.

Answer

f (x) =

-

\frac{1}{3} e^{x} -

2;

the domain is

(- \infty, \infty);

the range is

(- \infty, −2);

the horizontal asymptote is

y = −2.

6.2-1 §6.2 source ↗

What role does the horizontal asymptote of an exponential function play in telling us about the end behavior of the graph?

Answer

An asymptote is a line that the graph of a function approaches, as

x

either increases or decreases without bound. The horizontal asymptote of an exponential function tells us the limit of the function’s values as the independent variable gets either extremely large or extremely small.

6.2-3 §6.2 source ↗

The graph of

f (x) = 3^{x}

is reflected about the y-axis and stretched vertically by a factor of

4.

What is the equation of the new function,

g (x) ?

State its y-intercept, domain, and range.

Answer

g (x) =

4 {(3)}^{- x};

y-intercept:

(0, 4);

Domain: all real numbers; Range: all real numbers greater than

0.

6.2-5 §6.2 source ↗

The graph of

f (x) = 10^{x}

is reflected about the x-axis and shifted upward

7

units. What is the equation of the new function,

g (x) ?

State its y-intercept, domain, and range.

Answer

g (x) =

-

10^{x} +

7;

y-intercept:

(0, 6);

Domain: all real numbers; Range: all real numbers less than

7.

6.2-7 §6.2 source ↗

The graph of

f x =

- \frac{1}{2} {(\frac{1}{4})}^{x - 2} +

4

is shifted downward

4

units, and then shifted left

2

units, stretched vertically by a factor of

4,

and reflected about the x-axis. What is the equation of the new function,

g (x) ?

State its y-intercept, domain, and range.

Answer

g (x) =

2 {(\frac{1}{4})}^{x};

y-intercept:

(0, 2);

Domain: all real numbers; Range: all real numbers greater than

0.

6.2-9 §6.2 source ↗

For the following exercises, graph the function and its reflection about the y-axis on the same axes, and give the y-intercept.

$g (x) =$ $-$ $2 {(0.25)}^{x}$

Answer

y-intercept: $(0, - 2)$

6.2-11 §6.2 source ↗

For the following exercises, graph each set of functions on the same axes.

$f (x) =$ $3 {(\frac{1}{4})}^{x},$ $g (x) = 3 {(2)}^{x},$ and $h (x) = 3 {(4)}^{x}$

Answer

Graph of three functions, g(x)=3(2)^(x) in blue, h(x)=3(4)^(x) in green, and f(x)=3(1/4)^(x) in orange.

6.2-13 §6.2 source ↗

For the following exercises, match each function with one of the graphs in Figure 12.

Figure 12

$f (x) =$ $2 {(0.69)}^{x}$

Answer

B

6.2-15 §6.2 source ↗

For the following exercises, match each function with one of the graphs in Figure 12.

Figure 12

$f (x) =$ $2 {(0.81)}^{x}$

Answer

A

6.2-17 §6.2 source ↗

For the following exercises, match each function with one of the graphs in Figure 12.

Figure 12

$f (x) =$ $2 {(1.59)}^{x}$

Answer

E

6.2-19 §6.2 source ↗

For the following exercises, use the graphs shown in Figure 13. All have the form $f (x) = a b^{x} .$

Figure 13

Which graph has the largest value for $b ?$

Answer

D

6.2-21 §6.2 source ↗

For the following exercises, use the graphs shown in Figure 13. All have the form $f (x) = a b^{x} .$

Figure 13

Which graph has the largest value for $a ?$

Answer

C

6.2-23 §6.2 source ↗

For the following exercises, graph the function and its reflection about the x-axis on the same axes.

$f (x) = \frac{1}{2} {(4)}^{x}$

Answer

Graph of two functions, f(x)=(1/2)(4)^(x) in blue and -f(x)=(-1/2)(4)^x in orange.

6.2-25 §6.2 source ↗

For the following exercises, graph the function and its reflection about the x-axis on the same axes.

$f (x) =$ $-$ $4 {(2)}^{x} +$ $2$

Answer

6.2-27 §6.2 source ↗

For the following exercises, graph the transformation of $f (x) = 2^{x} .$ Give the horizontal asymptote, the domain, and the range.

$h (x) = 2^{x} + 3$

Answer

Horizontal asymptote: $h (x) = 3;$ Domain: all real numbers; Range: all real numbers strictly greater than $3.$

6.2-29 §6.2 source ↗

For the following exercises, describe the end behavior of the graphs of the functions.

$f (x) =$ $-$ $5 {(4)}^{x} -$ $1$

Answer

As

x \to \infty

,

f (x) \to - \infty

;
As

x \to - \infty

,

f (x) \to - 1

6.2-31 §6.2 source ↗

For the following exercises, describe the end behavior of the graphs of the functions.

$f (x) =$ $3 {(4)}^{- x} +$ $2$

Answer

As

x \to \infty

,

f (x) \to 2

;
As

x \to - \infty

,

f (x) \to \infty

6.2-33 §6.2 source ↗

For the following exercises, start with the graph of $f (x) = 4^{x} .$ Then write a function that results from the given transformation.

Shift $f (x)$ 3 units downward

Answer

f (x) = 4^{x} - 3

6.2-35 §6.2 source ↗

For the following exercises, start with the graph of $f (x) = 4^{x} .$ Then write a function that results from the given transformation.

Shift $f (x)$ 5 units right

Answer

f (x) = 4^{x - 5}

6.2-37 §6.2 source ↗

For the following exercises, start with the graph of $f (x) = 4^{x} .$ Then write a function that results from the given transformation.

Reflect $f (x)$ about the y-axis

Answer

f (x) = 4^{- x}

6.2-39 §6.2 source ↗

For the following exercises, each graph is a transformation of $y = 2^{x} .$ Write an equation describing the transformation.

Graph of f(x)=2^(x) with the following translations: a reflection about the x-axis, and a shift up by 3.

Answer

y = - 2^{x} + 3

6.2-41 §6.2 source ↗

For the following exercises, find an exponential equation for the graph.

Graph of f(x)=3^(x) with the following translations: vertical stretch of 2, a reflection about the x-axis, and a shift up by 7.

Answer

y = - 2 {(3)}^{x} + 7

6.2-43 §6.2 source ↗

For the following exercises, evaluate the exponential functions for the indicated value of $x .$

$g (x) =$ $\frac{1}{3} {(7)}^{x - 2}$ for $g (6) .$

Answer

g (6) =

800 +

\frac{1}{3} \approx

800.3333

6.2-45 §6.2 source ↗

For the following exercises, evaluate the exponential functions for the indicated value of $x .$

$h (x) =$ $-$ $\frac{1}{2} {(\frac{1}{2})}^{x} +$ $6$ for $h (- 7) .$

Answer

h (- 7) = - 58

6.2-47 §6.2 source ↗

For the following exercises, use a graphing calculator to approximate the solutions of the equation. Round to the nearest thousandth.

$116 = \frac{1}{4} {(\frac{1}{8})}^{x}$

Answer

x \approx - 2.953

6.2-49 §6.2 source ↗

For the following exercises, use a graphing calculator to approximate the solutions of the equation. Round to the nearest thousandth.

$5 =$ $3 {(\frac{1}{2})}^{x - 1} -$ $2$

Answer

x \approx - 0.222

6.2-51 §6.2 source ↗

For the following exercises, use a graphing calculator to approximate the solutions of the equation. Round to the nearest thousandth.

Explore and discuss the graphs of $F (x) = {(b)}^{x}$ and $G (x) = {(\frac{1}{b})}^{x} .$ Then make a conjecture about the relationship between the graphs of the functions $b^{x}$ and ${(\frac{1}{b})}^{x}$ for any real number $b > 0.$

Answer

The graph of

G (x) = {(\frac{1}{b})}^{x}

is the refelction about the y-axis of the graph of

F (x) = b^{x};

For any real number

b > 0

and function

f (x) = b^{x},

the graph of

{(\frac{1}{b})}^{x}

is the the reflection about the y-axis,

F (- x) .

6.2-53 §6.2 source ↗

For the following exercises, use a graphing calculator to approximate the solutions of the equation. Round to the nearest thousandth.

Explore and discuss the graphs of $f (x) = 4^{x},$ $g (x) = 4^{x - 2},$ and $h (x) =$ $(\frac{1}{16}) 4^{x} .$ Then make a conjecture about the relationship between the graphs of the functions $b^{x}$ and $(\frac{1}{b^{n}}) b^{x}$ for any real number n and real number $b > 0.$

Answer

The graphs of

g (x)

and

h (x)

are the same and are a horizontal shift to the right of the graph of

f (x);

For any real number n, real number

b > 0,

and function

f (x) = b^{x},

the graph of

(\frac{1}{b^{n}}) b^{x}

is the horizontal shift

f (x - n) .

Try It #1 §6.3 source ↗

Write the following logarithmic equations in exponential form.

ⓐ $\log_{10} (1, 000, 000) =$ $6$

ⓑ $\log_{5} (25) = 2$

Answer

ⓐ $\log_{10} (1, 000, 000) =$ $6$ is equivalent to $10^{6} =$ $1,$ $000,$ $000$

ⓑ $\log_{5} (25) = 2$ is equivalent to $5^{2} = 25$

Try It #2 §6.3 source ↗

Write the following exponential equations in logarithmic form.

ⓐ $3^{2} = 9$

ⓑ $5^{3} = 125$

Answer

ⓐ $3^{2} = 9$ is equivalent to $\log_{3} (9) = 2$

ⓑ $5^{3} = 125$ is equivalent to $\log_{5} (125) = 3$

Try It #3 §6.3 source ↗

Solve

y =

\log_{121} (11)

without using a calculator.

Answer

\log_{121} (11) =

\frac{1}{2}

(recalling that

\sqrt{121} =

{(121)}^{\frac{1}{2}} =

11

)

Try It #4 §6.3 source ↗

Evaluate

y = \log_{2} (\frac{1}{32})

without using a calculator.

Answer

\log_{2} (\frac{1}{32}) =

-

5

Try It #5 §6.3 source ↗

Evaluate

y =

\log (1, 000, 000) .

Answer

\log (1, 000, 000) =

6

Try It #6 §6.3 source ↗

Evaluate

y = \log (123)

to four decimal places using a calculator.

Answer

\log (123) \approx

2.0899

Try It #7 §6.3 source ↗

The amount of energy released from one earthquake was

8,500

times greater than the amount of energy released from another. The equation

10^{x} = 8500

represents this situation, where

x

is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes?

Answer

The difference in magnitudes was about

3.929.

Try It #8 §6.3 source ↗

Evaluate

\ln (−500) .

Answer

It is not possible to take the logarithm of a negative number in the set of real numbers.

6.3-1 §6.3 source ↗

What is a base

b

logarithm? Discuss the meaning by interpreting each part of the equivalent equations

b^{y} = x

and

\log_{b} x = y

for

b > 0, b \neq 1.

Answer

A logarithm is an exponent. Specifically, it is the exponent to which a base

b

is raised to produce a given value. In the expressions given, the base

b

has the same value. The exponent,

y,

in the expression

b^{y}

can also be written as the logarithm,

\log_{b} x,

and the value of

x

is the result of raising

b

to the power of

y .

6.3-3 §6.3 source ↗

How can the logarithmic equation

\log_{b} x = y

be solved for

x

using the properties of exponents?

Answer

Since the equation of a logarithm is equivalent to an exponential equation, the logarithm can be converted to the exponential equation

b^{y} = x,

and then properties of exponents can be applied to solve for

x .

6.3-5 §6.3 source ↗

Discuss the meaning of the natural logarithm. What is its relationship to a logarithm with base

b,

and how does the notation differ?

Answer

The natural logarithm is a special case of the logarithm with base

b

in that the natural log always has base

e .

Rather than notating the natural logarithm as

\log_{e} (x),

the notation used is

\ln (x) .

6.3-7 §6.3 source ↗

For the following exercises, rewrite each equation in exponential form.

${log}_{a} (b) = c$

Answer

a^{c} = b

6.3-9 §6.3 source ↗

For the following exercises, rewrite each equation in exponential form.

$\log_{x} (64) = y$

Answer

x^{y} = 64

6.3-11 §6.3 source ↗

For the following exercises, rewrite each equation in exponential form.

$\log_{15} (a) = b$

Answer

15^{b} = a

6.3-13 §6.3 source ↗

For the following exercises, rewrite each equation in exponential form.

$\log_{13} (142) =$ $a$

Answer

13^{a} = 142

6.3-15 §6.3 source ↗

For the following exercises, rewrite each equation in exponential form.

$ln (w) = n$

Answer

e^{n} = w

6.3-17 §6.3 source ↗

For the following exercises, rewrite each equation in logarithmic form.

$c^{d} = k$

Answer

{log}_{c} (k) = d

6.3-19 §6.3 source ↗

For the following exercises, rewrite each equation in logarithmic form.

$19^{x} = y$

Answer

\log_{19} y = x

6.3-21 §6.3 source ↗

For the following exercises, rewrite each equation in logarithmic form.

$n^{4} = 103$

Answer

\log_{n} (103) = 4

6.3-23 §6.3 source ↗

For the following exercises, rewrite each equation in logarithmic form.

$y^{x} = \frac{39}{100}$

Answer

\log_{y} (\frac{39}{100}) =

x

6.3-25 §6.3 source ↗

For the following exercises, rewrite each equation in logarithmic form.

$e^{k} = h$

Answer

ln (h) = k

6.3-27 §6.3 source ↗

For the following exercises, solve for $x$ by converting the logarithmic equation to exponential form.

${log}_{2} (x) = - 3$

Answer

x = 2^{- 3} = \frac{1}{8}

6.3-29 §6.3 source ↗

For the following exercises, solve for $x$ by converting the logarithmic equation to exponential form.

$\log_{3} (x) = 3$

Answer

x = 3^{3} = 27

6.3-31 §6.3 source ↗

For the following exercises, solve for $x$ by converting the logarithmic equation to exponential form.

${log}_{9} (x) = \frac{1}{2}$

Answer

x = 9^{\frac{1}{2}} = 3

6.3-33 §6.3 source ↗

For the following exercises, solve for $x$ by converting the logarithmic equation to exponential form.

$\log_{6} (x) = - 3$

Answer

x = 6^{- 3} = \frac{1}{216}

6.3-35 §6.3 source ↗

For the following exercises, solve for $x$ by converting the logarithmic equation to exponential form.

$ln (x) = 2$

Answer

x = e^{2}

6.3-37 §6.3 source ↗

For the following exercises, use the definition of common and natural logarithms to simplify.

$10^{log (32)}$

Answer

32

6.3-39 §6.3 source ↗

For the following exercises, use the definition of common and natural logarithms to simplify.

$e^{\ln (1.06)}$

Answer

1.06

6.3-41 §6.3 source ↗

For the following exercises, use the definition of common and natural logarithms to simplify.

$e^{\ln (10.125)} +$ $4$

Answer

14.125

6.3-43 §6.3 source ↗

For the following exercises, evaluate the base $b$ logarithmic expression without using a calculator.

${log}_{6} (\sqrt{6})$

Answer

\frac{1}{2}

6.3-45 §6.3 source ↗

For the following exercises, evaluate the base $b$ logarithmic expression without using a calculator.

$6 {log}_{8} (4)$

Answer

4

6.3-47 §6.3 source ↗

For the following exercises, evaluate the common logarithmic expression without using a calculator.

$log (0.001)$

Answer

- 3

6.3-49 §6.3 source ↗

For the following exercises, evaluate the common logarithmic expression without using a calculator.

$2 log (100^{- 3})$

Answer

- 12

6.3-51 §6.3 source ↗

For the following exercises, evaluate the natural logarithmic expression without using a calculator.

$ln (1)$

Answer

0

6.3-53 §6.3 source ↗

For the following exercises, evaluate the natural logarithmic expression without using a calculator.

$25 ln (e^{\frac{2}{5}})$

Answer

10

6.3-55 §6.3 source ↗

For the following exercises, evaluate each expression using a calculator. Round to the nearest thousandth.

$ln (15)$

Answer

2.708

6.3-57 §6.3 source ↗

For the following exercises, evaluate each expression using a calculator. Round to the nearest thousandth.

$log (\sqrt{2})$

Answer

0.151

6.3-59 §6.3 source ↗

For the following exercises, evaluate each expression using a calculator. Round to the nearest thousandth.

Is $x = 0$ in the domain of the function $f (x) =$ $\log (x) ?$ If so, what is the value of the function when $x = 0 ?$ Verify the result.

Answer

No, the function has no defined value for

x = 0.

To verify, suppose

x = 0

is in the domain of the function

f (x) =

\log (x) .

Then there is some number

n

such that

n = \log (0) .

Rewriting as an exponential equation gives:

10^{n} = 0,

which is impossible since no such real number

n

exists. Therefore,

x = 0

is not the domain of the function

f (x) =

\log (x) .

6.3-61 §6.3 source ↗

For the following exercises, evaluate each expression using a calculator. Round to the nearest thousandth.

Is there a number $x$ such that $\ln x = 2 ?$ If so, what is that number? Verify the result.

Answer

Yes. Suppose there exists a real number

x

such that

\ln x = 2.

Rewriting as an exponential equation gives

x = e^{2},

which is a real number. To verify, let

x = e^{2} .

Then, by definition,

\ln (x) =

\ln (e^{2}) =

2.

6.3-63 §6.3 source ↗

For the following exercises, evaluate each expression using a calculator. Round to the nearest thousandth.

Is the following true: $\frac{\ln (e^{1.725})}{\ln (1)} =$ $1.725 ?$ Verify the result.

Answer

No;

\ln (1) = 0,

so

\frac{\ln (e^{1.725})}{\ln (1)}

is undefined.

6.3-65 §6.3 source ↗

For the following exercises, evaluate each expression using a calculator. Round to the nearest thousandth.

Refer to the previous exercise. Suppose the light meter on a camera indicates an $E I$ of $- 2,$ and the desired exposure time is 16 seconds. What should the f-stop setting be?

Answer

2

Try It #1 §6.4 source ↗

What is the domain of

f (x) =

\log_{5} (x - 2) +

1 ?

Answer

(2, \infty)

Try It #2 §6.4 source ↗

What is the domain of

f (x) =

\log (x - 5) +

2 ?

Answer

(5, \infty)

Try It #3 §6.4 source ↗

Graph

f (x) =

\log_{\frac{1}{5}} (x) .

State the domain, range, and asymptote.

Answer

Graph of f(x)=log_(1/5)(x) with labeled points at (1/5, 1) and (1, 0). The y-axis is the asymptote.

The domain is $(0, \infty),$ the range is $(- \infty, \infty),$ and the vertical asymptote is $x = 0.$

Try It #4 §6.4 source ↗

Sketch a graph of

f (x) =

\log_{3} (x + 4)

alongside its parent function. Include the key points and asymptotes on the graph. State the domain, range, and asymptote.

Answer

Graph of two functions. The parent function is y=log_3(x), with an asymptote at x=0 and labeled points at (1, 0), and (3, 1).The translation function f(x)=log_3(x+4) has an asymptote at x=-4 and labeled points at (-3, 0) and (-1, 1).

The domain is $(- 4, \infty),$ the range $(- \infty, \infty),$ and the asymptote $x = - 4.$

Try It #5 §6.4 source ↗

Sketch a graph of

f (x) =

\log_{2} (x) +

2

alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.

Answer

Graph of two functions. The parent function is y=log_2(x), with an asymptote at x=0 and labeled points at (1, 0), and (2, 1).The translation function f(x)=log_2(x)+2 has an asymptote at x=0 and labeled points at (0.25, 0) and (0.5, 1).

The domain is $(0, \infty),$ the range is $(- \infty, \infty),$ and the vertical asymptote is $x = 0.$

Try It #6 §6.4 source ↗

Sketch a graph of

f (x) =

\frac{1}{2} \log_{4} (x)

alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.

Answer

Graph of two functions. The parent function is y=log_4(x), with an asymptote at x=0 and labeled points at (1, 0), and (4, 1).The translation function f(x)=(1/2)log_4(x) has an asymptote at x=0 and labeled points at (1, 0) and (16, 1).

The domain is $(0, \infty),$ the range is $(- \infty, \infty),$ and the vertical asymptote is $x = 0.$

Try It #7 §6.4 source ↗

Sketch a graph of the function

f (x) =

3 \log (x - 2) +

1.

State the domain, range, and asymptote.

Answer

Graph of f(x)=3log(x-2)+1 with an asymptote at x=2.

The domain is $(2, \infty),$ the range is $(- \infty, \infty),$ and the vertical asymptote is $x = 2.$

Try It #8 §6.4 source ↗

Graph

f (x) =

-

\log (- x) .

State the domain, range, and asymptote.

Answer

Graph of f(x)=-log(-x) with an asymptote at x=0.

The domain is $(- \infty, 0),$ the range is $(- \infty, \infty),$ and the vertical asymptote is $x = 0.$

Try It #9 §6.4 source ↗

Solve

5 \log (x + 2) =

4 -

\log (x)

graphically. Round to the nearest thousandth.

Answer

x \approx 3.049

Try It #10 §6.4 source ↗

What is the vertical asymptote of

f (x) =

3 +

\ln (x - 1) ?

Answer

x = 1

Try It #11 §6.4 source ↗

Give the equation of the natural logarithm graphed in Figure 16.

Graph of a logarithmic function with a vertical asymptote at x=-3, has been vertically stretched by 2, and passes through the points (-1, -1).

Figure 16

Answer

f (x) =

2 \ln (x + 3) -

1

6.4-1 §6.4 source ↗

The inverse of every logarithmic function is an exponential function and vice-versa. What does this tell us about the relationship between the coordinates of the points on the graphs of each?

Answer

Since the functions are inverses, their graphs are mirror images about the line

y = x .

So for every point

(a, b)

on the graph of a logarithmic function, there is a corresponding point

(b, a)

on the graph of its inverse exponential function.

6.4-3 §6.4 source ↗

What type(s) of translation(s), if any, affect the domain of a logarithmic function?

Answer

Shifting the function right or left and reflecting the function about the y-axis will affect its domain.

6.4-5 §6.4 source ↗

Does the graph of a general logarithmic function have a horizontal asymptote? Explain.

Answer

No. A horizontal asymptote would suggest a limit on the range, and the range of any logarithmic function in general form is all real numbers.

6.4-7 §6.4 source ↗

For the following exercises, state the domain and range of the function.

$h (x) =$ $\ln (\frac{1}{2} - x)$

Answer

Domain:

(- \infty, \frac{1}{2});

Range:

(- \infty, \infty)

6.4-9 §6.4 source ↗

For the following exercises, state the domain and range of the function.

$h (x) =$ $\ln (4 x + 17) -$ $5$

Answer

Domain:

(- \frac{17}{4}, \infty);

Range:

(- \infty, \infty)

6.4-11 §6.4 source ↗

For the following exercises, state the domain and the vertical asymptote of the function.

$f (x) =$ $\log_{b} (x - 5)$

Answer

Domain:

(5, \infty);

Vertical asymptote:

x = 5

6.4-13 §6.4 source ↗

For the following exercises, state the domain and the vertical asymptote of the function.

$f (x) =$ $\log (3 x + 1)$

Answer

Domain:

(- \frac{1}{3}, \infty);

Vertical asymptote:

x = - \frac{1}{3}

6.4-15 §6.4 source ↗

For the following exercises, state the domain and the vertical asymptote of the function.

$g (x) =$ $-$ $\ln (3 x + 9) -$ $7$

Answer

Domain:

(- 3, \infty);

Vertical asymptote:

x = - 3

6.4-17 §6.4 source ↗

For the following exercises, state the domain, vertical asymptote, and end behavior of the function.

$f (x) =$ $\log (x - \frac{3}{7})$

Answer

Domain:

(\frac{3}{7}, \infty)

;
Vertical asymptote:

x = \frac{3}{7}

; End behavior: as

x \to

{(\frac{3}{7})}^{+},

f (x) \to

-

\infty

and as

x \to \infty, f (x) \to \infty

6.4-19 §6.4 source ↗

For the following exercises, state the domain, vertical asymptote, and end behavior of the function.

$g (x) =$ $\ln (2 x + 6) -$ $5$

Answer

Domain:

(- 3, \infty)

; Vertical asymptote:

x = - 3

;
End behavior: as

x \to - 3^{+}

,

f (x) \to - \infty

and as

x \to \infty

,

f (x) \to \infty

6.4-21 §6.4 source ↗

For the following exercises, state the domain, range, and x- and y-intercepts, if they exist. If they do not exist, write DNE.

$h (x) =$ $\log_{4} (x - 1) +$ $1$

Answer

Domain:

(1, \infty);

Range:

(- \infty, \infty);

Vertical asymptote:

x = 1;

x-intercept:

(\frac{5}{4}, 0);

y-intercept: DNE

6.4-23 §6.4 source ↗

For the following exercises, state the domain, range, and x- and y-intercepts, if they exist. If they do not exist, write DNE.

$g (x) =$ $\ln (- x) -$ $2$

Answer

Domain:

(- \infty, 0);

Range:

(- \infty, \infty);

Vertical asymptote:

x = 0;

x-intercept:

(- e^{2}, 0);

y-intercept: DNE

6.4-25 §6.4 source ↗

For the following exercises, state the domain, range, and x- and y-intercepts, if they exist. If they do not exist, write DNE.

$h (x) =$ $3 \ln (x) -$ $9$

Answer

Domain:

(0, \infty);

Range:

(- \infty, \infty);

Vertical asymptote:

x = 0;

x-intercept:

(e^{3}, 0);

y-intercept: DNE

6.4-27 §6.4 source ↗

For the following exercises, match each function in Figure 17 with the letter corresponding to its graph.

Figure 17

$f (x) = \ln (x)$

Answer

B

6.4-29 §6.4 source ↗

For the following exercises, match each function in Figure 17 with the letter corresponding to its graph.

Figure 17

$h (x) =$ $\log_{5} (x)$

Answer

C

6.4-31 §6.4 source ↗

For the following exercises, match each function in Figure 18 with the letter corresponding to its graph.

Figure 18

$f (x) =$ $\log_{\frac{1}{3}} (x)$

Answer

B

6.4-33 §6.4 source ↗

For the following exercises, match each function in Figure 18 with the letter corresponding to its graph.

Figure 18

$h (x) =$ $\log_{\frac{3}{4}} (x)$

Answer

C

6.4-35 §6.4 source ↗

For the following exercises, sketch the graphs of each pair of functions on the same axis.

$f (x) = \log (x)$ and $g (x) =$ $\log_{\frac{1}{2}} (x)$

Answer

Graph of two functions, g(x) = log_(1/2)(x) in orange and f(x)=log(x) in blue.

6.4-37 §6.4 source ↗

For the following exercises, sketch the graphs of each pair of functions on the same axis.

$f (x) = e^{x}$ and $g (x) = \ln (x)$

Answer

Graph of two functions, g(x) = ln(1/2)(x) in orange and f(x)=e^(x) in blue.

6.4-39 §6.4 source ↗

For the following exercises, match each function in Figure 19 with the letter corresponding to its graph.

Figure 19

$g (x) =$ $-$ $\log_{4} (x + 2)$

Answer

C

6.4-41 §6.4 source ↗

For the following exercises, sketch the graph of the indicated function.

$f (x) =$ $\log_{2} (x + 2)$

Answer

6.4-43 §6.4 source ↗

For the following exercises, sketch the graph of the indicated function.

$f (x) = \ln (- x)$

Answer

6.4-45 §6.4 source ↗

For the following exercises, sketch the graph of the indicated function.

$g (x) =$ $\log (6 - 3 x) +$ $1$

Answer

6.4-47 §6.4 source ↗

For the following exercises, write a logarithmic equation corresponding to the graph shown.

Use $y = \log_{2} (x)$ as the parent function.

The graph y=log_2(x) has been reflected over the y-axis and shifted to the right by 1.

Answer

f (x) =

\log_{2} (- (x - 1))

6.4-49 §6.4 source ↗

For the following exercises, write a logarithmic equation corresponding to the graph shown.

Use $f (x) =$ $\log_{4} (x)$ as the parent function.

The graph y=log_4(x) has been vertically stretched by 3, and shifted to the left by 2.

Answer

f (x) =

3 \log_{4} (x + 2)

6.4-51 §6.4 source ↗

For the following exercises, use a graphing calculator to find approximate solutions to each equation.

$\log (x - 1) +$ $2 =$ $\ln (x - 1) +$ $2$

Answer

x = 2

6.4-53 §6.4 source ↗

For the following exercises, use a graphing calculator to find approximate solutions to each equation.

$\ln (x - 2) =$ $-$ $\ln (x + 1)$

Answer

x \approx 2 .303

6.4-55 §6.4 source ↗

For the following exercises, use a graphing calculator to find approximate solutions to each equation.

$\frac{1}{3} \log (1 - x) =$ $\log (x + 1) +$ $\frac{1}{3}$

Answer

x \approx - 0.472

6.4-57 §6.4 source ↗

For the following exercises, use a graphing calculator to find approximate solutions to each equation.

Explore and discuss the graphs of $f (x) =$ $\log_{\frac{1}{2}} (x)$ and $g (x) =$ $-$ $\log_{2} (x) .$ Make a conjecture based on the result.

Answer

The graphs of

f (x) =

\log_{\frac{1}{2}} (x)

and

g (x) =

-

\log_{2} (x)

appear to be the same; Conjecture: for any positive base

b \neq 1,

\log_{b} (x) =

-

\log_{\frac{1}{b}} (x) .

6.4-59 §6.4 source ↗

For the following exercises, use a graphing calculator to find approximate solutions to each equation.

What is the domain of the function $f (x) =$ $\ln (\frac{x + 2}{x - 4}) ?$ Discuss the result.

Answer

Recall that the argument of a logarithmic function must be positive, so we determine where $\frac{x + 2}{x - 4} > 0$ . From the graph of the function $f (x) =$ $\frac{x + 2}{x - 4},$ note that the graph lies above the x-axis on the interval $(- \infty, - 2)$ and again to the right of the vertical asymptote, that is $(4, \infty) .$ Therefore, the domain is $(- \infty, - 2) \cup (4, \infty) .$

A graph of a rational function is shown on an xy-coordinate plane. The x-axis ranges from -10 to 10, and the y-axis ranges from -10 to 10. A vertical dashed orange line, labeled x = 4, represents a vertical asymptote. The graph consists of two smooth blue curves. The left curve extends from negative infinity in the second quadrant, passes through the x-intercept at (-2, 0) (marked with a blue dot), and decreases towards negative infinity as it approaches the vertical asymptote x=4 from the left. The right curve begins at positive infinity as it approaches the vertical asymptote x=4 from the right, then decreases and approaches the x-axis from above as x tends towards positive infinity in the first quadrant.

Try It #1 §6.5 source ↗

Expand

\log_{b} (8 k) .

Answer

\log_{b} 2 +

\log_{b} 2 +

\log_{b} 2 +

\log_{b} k =

3 \log_{b} 2 +

\log_{b} k

Try It #2 §6.5 source ↗

Expand

\log_{3} (\frac{7 x^{2} + 21 x}{7 x (x - 1) (x - 2)}) .

Answer

\log_{3} (x + 3) -

\log_{3} (x - 1) -

\log_{3} (x - 2)

Try It #3 §6.5 source ↗

Expand

\ln x^{2} .

Answer

2 \ln x

Try It #4 §6.5 source ↗

Expand

\ln (\frac{1}{x^{2}}) .

Answer

- 2 \ln (x)

Try It #5 §6.5 source ↗

Rewrite

2 \log_{3} 4

using the power rule for logs to a single logarithm with a leading coefficient of 1.

Answer

\log_{3} 16

Try It #6 §6.5 source ↗

Expand

\log (\frac{x^{2} y^{3}}{z^{4}}) .

Answer

2 \log x +

3 \log y -

4 \log z

Try It #7 §6.5 source ↗

Expand

\ln (\sqrt[3]{x^{2}}) .

Answer

\frac{2}{3} \ln x

Try It #8 §6.5 source ↗

Expand

\ln (\frac{\sqrt{(x - 1) {(2 x + 1)}^{2}}}{(x^{2} - 9)}) .

Answer

\frac{1}{2} \ln (x - 1) +

\ln (2 x + 1) -

\ln (x + 3) -

\ln (x - 3)

Try It #9 §6.5 source ↗

Condense

\log 3 -

\log 4 +

\log 5 -

\log 6.

Answer

\log (\frac{3 \cdot 5}{4 \cdot 6});

can also be written

\log (\frac{5}{8})

by reducing the fraction to lowest terms.

Try It #10 §6.5 source ↗

Rewrite

\log (5) +

0.5 \log (x) -

\log (7 x - 1) +

3 \log (x - 1)

as a single logarithm.

Answer

\log (\frac{5 {(x - 1)}^{3} \sqrt{x}}{(7 x - 1)})

Try It #11 §6.5 source ↗

Condense

4 (3 \log (x) + \log (x + 5) - \log (2 x + 3)) .

Answer

\log \frac{x^{12} {(x + 5)}^{4}}{{(2 x + 3)}^{4}};

this answer could also be written

\log {(\frac{x^{3} (x + 5)}{(2 x + 3)})}^{4} .

Try It #12 §6.5 source ↗

How does the pH change when the concentration of positive hydrogen ions is decreased by half?

Answer

The pH increases by about 0.301.

Try It #13 §6.5 source ↗

Change

\log_{0.5} 8

to a quotient of natural logarithms.

Answer

\frac{\ln 8}{\ln 0.5}

Try It #14 §6.5 source ↗

Evaluate

\log_{5} (100)

using the change-of-base formula.

Answer

\frac{\ln 100}{\ln 5} \approx

\frac{4.6051}{1.6094} =

2.861

6.5-1 §6.5 source ↗

How does the power rule for logarithms help when solving logarithms with the form

\log_{b} (\sqrt[n]{x}) ?

Answer

Any root expression can be rewritten as an expression with a rational exponent so that the power rule can be applied, making the logarithm easier to calculate. Thus,

\log_{b} (x^{\frac{1}{n}}) =

\frac{1}{n} \log_{b} (x) .

6.5-3 §6.5 source ↗

For the following exercises, expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs.

$\log_{b} (7 x \cdot 2 y)$

Answer

\log_{b} (2) +

\log_{b} (7) +

\log_{b} (x) +

\log_{b} (y)

6.5-5 §6.5 source ↗

For the following exercises, expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs.

$\log_{b} (\frac{13}{17})$

Answer

\log_{b} (13) -

\log_{b} (17)

6.5-7 §6.5 source ↗

For the following exercises, expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs.

$\ln (\frac{1}{4^{k}})$

Answer

- k \ln (4)

6.5-9 §6.5 source ↗

For the following exercises, condense to a single logarithm if possible.

$\ln (7) +$ $\ln (x) +$ $\ln (y)$

Answer

\ln (7 x y)

6.5-11 §6.5 source ↗

For the following exercises, condense to a single logarithm if possible.

$\log_{b} (28) -$ $\log_{b} (7)$

Answer

\log_{b} (4)

6.5-13 §6.5 source ↗

For the following exercises, condense to a single logarithm if possible.

$- \log_{b} (\frac{1}{7})$

Answer

{log}_{b} (7)

6.5-15 §6.5 source ↗

For the following exercises, use the properties of logarithms to expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs.

$\log (\frac{x^{15} y^{13}}{z^{19}})$

Answer

15 \log (x) +

13 \log (y) -

19 \log (z)

6.5-17 §6.5 source ↗

For the following exercises, use the properties of logarithms to expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs.

$\log (\sqrt{x^{3} y^{- 4}})$

Answer

\frac{3}{2} \log (x) -

2 \log (y)

6.5-19 §6.5 source ↗

For the following exercises, use the properties of logarithms to expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs.

$\log (x^{2} y^{3} \sqrt[3]{x^{2} y^{5}})$

Answer

\frac{8}{3} \log (x) +

\frac{14}{3} \log (y)

6.5-21 §6.5 source ↗

For the following exercises, condense each expression to a single logarithm using the properties of logarithms.

$\ln (6 x^{9}) -$ $\ln (3 x^{2})$

Answer

\ln (2 x^{7})

6.5-23 §6.5 source ↗

For the following exercises, condense each expression to a single logarithm using the properties of logarithms.

$\log (x) -$ $\frac{1}{2} \log (y) +$ $3 \log (z)$

Answer

\log (\frac{x z^{3}}{\sqrt{y}})

6.5-25 §6.5 source ↗

For the following exercises, rewrite each expression as an equivalent ratio of logs using the indicated base.

$\log_{7} (15)$ to base $e$

Answer

\log_{7} (15) =

\frac{\ln (15)}{\ln (7)}

6.5-27 §6.5 source ↗

For the following exercises, suppose $\log_{5} (6) = a$ and $\log_{5} (11) = b .$ Use the change-of-base formula along with properties of logarithms to rewrite each expression in terms of $a$ and $b .$ Show the steps for solving.

$\log_{11} (5)$

Answer

\log_{11} (5) =

\frac{\log_{5} (5)}{\log_{5} (11)} =

\frac{1}{b}

6.5-29 §6.5 source ↗

For the following exercises, suppose $\log_{5} (6) = a$ and $\log_{5} (11) = b .$ Use the change-of-base formula along with properties of logarithms to rewrite each expression in terms of $a$ and $b .$ Show the steps for solving.

$\log_{11} (\frac{6}{11})$

Answer

\log_{11} (\frac{6}{11}) =

\frac{\log_{5} (\frac{6}{11})}{\log_{5} (11)} =

\frac{\log_{5} (6) - \log_{5} (11)}{\log_{5} (11)} =

\frac{a - b}{b} =

\frac{a}{b} -

1

6.5-31 §6.5 source ↗

For the following exercises, use properties of logarithms to evaluate without using a calculator.

$6 \log_{8} (2) +$ $\frac{\log_{8} (64)}{3 \log_{8} (4)}$

Answer

3

6.5-33 §6.5 source ↗

For the following exercises, use the change-of-base formula to evaluate each expression as a quotient of natural logs. Use a calculator to approximate each to five decimal places.

$\log_{3} (22)$

Answer

2.81359

6.5-35 §6.5 source ↗

For the following exercises, use the change-of-base formula to evaluate each expression as a quotient of natural logs. Use a calculator to approximate each to five decimal places.

$\log_{6} (5.38)$

Answer

0.93913

6.5-37 §6.5 source ↗

For the following exercises, use the change-of-base formula to evaluate each expression as a quotient of natural logs. Use a calculator to approximate each to five decimal places.

$\log_{\frac{1}{2}} (4.7)$

Answer

- 2.23266

6.5-39 §6.5 source ↗

For the following exercises, use the change-of-base formula to evaluate each expression as a quotient of natural logs. Use a calculator to approximate each to five decimal places.

Use the quotient rule for logarithms to find all $x$ values such that $\log_{6} (x + 2) -$ $\log_{6} (x - 3) =$ $1.$ Show the steps for solving.

Answer

$x = 4;$ By the quotient rule: $\log_{6} (x + 2) -$ $\log_{6} (x - 3) =$ $\log_{6} (\frac{x + 2}{x - 3}) =$ $1.$

Rewriting as an exponential equation and solving for $x :$

$\begin{array}{l} 6^{1} & = \frac{x + 2}{x - 3} (after multiplying both sides by (x - 3)) \\ 6 (x - 3) & = x + 2 (carry out distributive multiplication on left) \\ 6 x - 18 & = x + 2 (add 18 to both sides, subtract x from both sides) \\ 5 x & = 20 \\ x & = 4 \end{array}$

Checking, we find that $\log_{6} (4 + 2) -$ $\log_{6} (4 - 3) =$ $\log_{6} (6) -$ $\log_{6} (1)$ is defined, so $x = 4.$

6.5-41 §6.5 source ↗

For the following exercises, use the change-of-base formula to evaluate each expression as a quotient of natural logs. Use a calculator to approximate each to five decimal places.

Prove that $\log_{b} (n) =$ $\frac{1}{\log_{n} (b)}$ for any positive integers $b > 1$ and $n > 1.$

Answer

Let

b

and

n

be positive integers greater than

1.

Then, by the change-of-base formula,

\log_{b} (n) =

\frac{\log_{n} (n)}{\log_{n} (b)} =

\frac{1}{\log_{n} (b)} .

Try It #1 §6.6 source ↗

Solve

5^{2 x} = 5^{3 x + 2} .

Answer

x = - 2

Try It #2 §6.6 source ↗

Solve

5^{2 x} = 25^{3 x + 2} .

Answer

x = - 1

Try It #3 §6.6 source ↗

Solve

5^{x} = \sqrt{5} .

Answer

x = \frac{1}{2}

Try It #4 §6.6 source ↗

Solve

2^{x} = −100.

Answer

The equation has no solution.

Try It #5 §6.6 source ↗

Solve

2^{x} = 3^{x + 1} .

Answer

x = \frac{\ln 3}{\ln (\frac{2}{3})}

Try It #6 §6.6 source ↗

Solve

3 e^{0.5 t} = 11.

Answer

t = 2 \ln (\frac{11}{3})

or

\ln {(\frac{11}{3})}^{2}

Try It #7 §6.6 source ↗

Solve

3 + e^{2 t} = 7 e^{2 t} .

Answer

t =

\ln (\frac{1}{\sqrt{2}}) =

-

\frac{1}{2} \ln (2)

Try It #8 §6.6 source ↗

Solve

e^{2 x} = e^{x} + 2.

Answer

x = \ln 2

Try It #9 §6.6 source ↗

Solve

6 + \ln x = 10.

Answer

x = e^{4}

Try It #10 §6.6 source ↗

Solve

2 \ln (x + 1) =

10.

Answer

x = e^{5} - 1

Try It #11 §6.6 source ↗

Use a graphing calculator to estimate the approximate solution to the logarithmic equation

2^{x} = 1000

to 2 decimal places.

Answer

x \approx 9.97

Try It #12 §6.6 source ↗

Solve

\ln (x^{2}) = \ln 1.

Answer

x = 1

or

x = - 1

Try It #13 §6.6 source ↗

How long will it take before twenty percent of our 1000-gram sample of uranium-235 has decayed?

Answer

t =

703,

800,

000 \times \frac{\ln (0.8)}{\ln (0.5)} years \approx

226,

572,

993 years .

6.6-1 §6.6 source ↗

How can an exponential equation be solved?

Answer

Determine first if the equation can be rewritten so that each side uses the same base. If so, the exponents can be set equal to each other. If the equation cannot be rewritten so that each side uses the same base, then apply the logarithm to each side and use properties of logarithms to solve.

6.6-3 §6.6 source ↗

When can the one-to-one property of logarithms be used to solve an equation? When can it not be used?

Answer

The one-to-one property can be used if both sides of the equation can be rewritten as a single logarithm with the same base. If so, the arguments can be set equal to each other, and the resulting equation can be solved algebraically. The one-to-one property cannot be used when each side of the equation cannot be rewritten as a single logarithm with the same base.

6.6-5 §6.6 source ↗

For the following exercises, use like bases to solve the exponential equation.

$64 \cdot 4^{3 x} = 16$

Answer

x = - \frac{1}{3}

6.6-7 §6.6 source ↗

For the following exercises, use like bases to solve the exponential equation.

$2^{- 3 n} \cdot \frac{1}{4} =$ $2^{n + 2}$

Answer

n = - 1

6.6-9 §6.6 source ↗

For the following exercises, use like bases to solve the exponential equation.

$\frac{36^{3 b}}{36^{2 b}} =$ $216^{2 - b}$

Answer

b = \frac{6}{5}

6.6-11 §6.6 source ↗

For the following exercises, use logarithms to solve.

$9^{x - 10} = 1$

Answer

x = 10

6.6-13 §6.6 source ↗

For the following exercises, use logarithms to solve.

$e^{r + 10} -$ $10 =$ $−42$

Answer

No solution

6.6-15 §6.6 source ↗

For the following exercises, use logarithms to solve.

$-$ $8 \cdot 10^{p + 7} -$ $7 =$ $−24$

Answer

p =

\log (\frac{17}{8}) -

7

6.6-17 §6.6 source ↗

For the following exercises, use logarithms to solve.

$e^{- 3 k} + 6 = 44$

Answer

k = - \frac{\ln (38)}{3}

6.6-19 §6.6 source ↗

For the following exercises, use logarithms to solve.

$-$ $6 e^{9 x + 8} +$ $2 =$ $−74$

Answer

x =

\frac{\ln (\frac{38}{3}) - 8}{9}

6.6-21 §6.6 source ↗

For the following exercises, use logarithms to solve.

$e^{2 x} -$ $e^{x} -$ $132 =$ $0$

Answer

x = \ln 12

6.6-23 §6.6 source ↗

For the following exercises, use logarithms to solve.

$10 e^{8 x + 3} + 2 = 8$

Answer

x = \frac{\ln (\frac{3}{5}) - 3}{8}

6.6-25 §6.6 source ↗

For the following exercises, use logarithms to solve.

$8 e^{- 5 x - 2} -$ $4 =$ $−90$

Answer

no solution

6.6-27 §6.6 source ↗

For the following exercises, use logarithms to solve.

$e^{2 x} - e^{x} - 6 = 0$

Answer

x = \ln (3)

6.6-29 §6.6 source ↗

For the following exercises, use the definition of a logarithm to rewrite the equation as an exponential equation.

$\log (\frac{1}{100}) =$ $−2$

Answer

10^{- 2} = \frac{1}{100}

6.6-31 §6.6 source ↗

For the following exercises, use the definition of a logarithm to solve the equation.

$5 \log_{7} n = 10$

Answer

n = 49

6.6-33 §6.6 source ↗

For the following exercises, use the definition of a logarithm to solve the equation.

$4 +$ $\log_{2} (9 k) =$ $2$

Answer

k = \frac{1}{36}

6.6-35 §6.6 source ↗

For the following exercises, use the definition of a logarithm to solve the equation.

$10 -$ $4 \ln (9 - 8 x) =$ $6$

Answer

x = \frac{9 - e}{8}

6.6-37 §6.6 source ↗

For the following exercises, use the one-to-one property of logarithms to solve.

$\log_{13} (5 n - 2) =$ $\log_{13} (8 - 5 n)$

Answer

n = 1

6.6-39 §6.6 source ↗

For the following exercises, use the one-to-one property of logarithms to solve.

$\ln (- 3 x) =$ $\ln (x^{2} - 6 x)$

Answer

No solution

6.6-41 §6.6 source ↗

For the following exercises, use the one-to-one property of logarithms to solve.

$\ln (x - 2) -$ $\ln (x) =$ $\ln (54)$

Answer

No solution

6.6-43 §6.6 source ↗

For the following exercises, use the one-to-one property of logarithms to solve.

$\ln (x^{2} - 10) +$ $\ln (9) =$ $\ln (10)$

Answer

x = \pm \frac{10}{3}

6.6-45 §6.6 source ↗

For the following exercises, solve each equation for $x .$

$\ln (x) +$ $\ln (x - 3) =$ $\ln (7 x)$

Answer

x = 10

6.6-47 §6.6 source ↗

For the following exercises, solve each equation for $x .$

$\ln (7) +$ $\ln (2 - 4 x^{2}) =$ $\ln (14)$

Answer

x = 0

6.6-49 §6.6 source ↗

For the following exercises, solve each equation for $x .$

$\ln (3) -$ $\ln (3 - 3 x) =$ $\ln (4)$

Answer

x = \frac{3}{4}

6.6-51 §6.6 source ↗

For the following exercises, solve the equation for $x,$ if there is a solution. Then graph both sides of the equation, and observe the point of intersection (if it exists) to verify the solution.

$\log_{9} (x) -$ $5 =$ $−4$

Answer

$x = 9$

6.6-53 §6.6 source ↗

For the following exercises, solve the equation for $x,$ if there is a solution. Then graph both sides of the equation, and observe the point of intersection (if it exists) to verify the solution.

$\ln (3 x) = 2$

Answer

$x = \frac{e^{2}}{3} \approx 2.5$

6.6-55 §6.6 source ↗

For the following exercises, solve the equation for $x,$ if there is a solution. Then graph both sides of the equation, and observe the point of intersection (if it exists) to verify the solution.

$\log (4) +$ $\log (- 5 x) =$ $2$

Answer

$x = - 5$

6.6-57 §6.6 source ↗

For the following exercises, solve the equation for $x,$ if there is a solution. Then graph both sides of the equation, and observe the point of intersection (if it exists) to verify the solution.

$\ln (4 x - 10) -$ $6 =$ $-$ $5$

Answer

$x = \frac{e + 10}{4} \approx 3.2$

6.6-59 §6.6 source ↗

For the following exercises, solve the equation for $x,$ if there is a solution. Then graph both sides of the equation, and observe the point of intersection (if it exists) to verify the solution.

$\log_{11} (- 2 x^{2} - 7 x) =$ $\log_{11} (x - 2)$

Answer

No solution

Graph of log_11(-2x^2-7x)=y and y=log_11(x-2).

6.6-61 §6.6 source ↗

For the following exercises, solve the equation for $x,$ if there is a solution. Then graph both sides of the equation, and observe the point of intersection (if it exists) to verify the solution.

$\log_{9} (3 - x) =$ $\log_{9} (4 x - 8)$

Answer

$x = \frac{11}{5} \approx 2.2$

Graph of log_9(3-x)=y and y=log_9(4x-8).

6.6-63 §6.6 source ↗

For the following exercises, solve the equation for $x,$ if there is a solution. Then graph both sides of the equation, and observe the point of intersection (if it exists) to verify the solution.

$\frac{3}{\log_{2} (10)} -$ $\log (x - 9) =$ $\log (44)$

Answer

$x = \frac{101}{11} \approx 9.2$

Graph of 3/log_2(10)-log(x-9)=y and y=log(44).

6.6-65 §6.6 source ↗

For the following exercises, solve for the indicated value, and graph the situation showing the solution point.

An account with an initial deposit of $$6,500$ earns $7.25 %$ annual interest, compounded continuously. How much will the account be worth after 20 years?

Answer

about $$ 27, 710.24$

Graph of f(x)=6500e^(0.0725x) with the labeled point at (20, 27710.24).

6.6-67 §6.6 source ↗

For the following exercises, solve for the indicated value, and graph the situation showing the solution point.

The population of a small town is modeled by the equation $P = 1650 e^{0.5 t}$ where $t$ is measured in years. In approximately how many years will the town’s population reach $20,000?$

Answer

about 5 years

Graph of P(t)=1650e^(0.5x) with the labeled point at (5, 20000).

6.6-69 §6.6 source ↗

For the following exercises, solve each equation by rewriting the exponential expression using the indicated logarithm. Then use a calculator to approximate the variable to 3 decimal places.

$e^{5 x} = 17$ using the natural log

Answer

\frac{\ln (17)}{5} \approx

0.567

6.6-71 §6.6 source ↗

For the following exercises, solve each equation by rewriting the exponential expression using the indicated logarithm. Then use a calculator to approximate the variable to 3 decimal places.

$3^{4 x - 5} = 38$ using the common log

Answer

x =

\frac{\log (38) + 5 \log (3)}{4 \log (3)} \approx

2.078

6.6-73 §6.6 source ↗

For the following exercises, use a calculator to solve the equation. Unless indicated otherwise, round all answers to the nearest ten-thousandth.

$7 e^{3 x - 5} +$ $7.9 =$ $47$

Answer

x \approx 2.2401

6.6-75 §6.6 source ↗

For the following exercises, use a calculator to solve the equation. Unless indicated otherwise, round all answers to the nearest ten-thousandth.

$\log (- 0.7 x - 9) =$ $1 +$ $5 \log (5)$

Answer

x \approx

-

44655.7143

6.6-77 §6.6 source ↗

For the following exercises, use a calculator to solve the equation. Unless indicated otherwise, round all answers to the nearest ten-thousandth.

The magnitude M of an earthquake is represented by the equation $M =$ $\frac{2}{3} \log (\frac{E}{E_{0}})$ where $E$ is the amount of energy released by the earthquake in joules and $E_{0} = 10^{4.4}$ is the assigned minimal measure released by an earthquake. To the nearest hundredth, what would the magnitude be of an earthquake releasing $1.4 \cdot 10^{13}$ joules of energy?

Answer

about

5.83

6.6-79 §6.6 source ↗

For the following exercises, use a calculator to solve the equation. Unless indicated otherwise, round all answers to the nearest ten-thousandth.

Recall the formula for continually compounding interest, $y = A e^{k t} .$ Use the definition of a logarithm along with properties of logarithms to solve the formula for time $t$ such that $t$ is equal to a single logarithm.

Answer

t =

\ln ({(\frac{y}{A})}^{\frac{1}{k}})

6.6-81 §6.6 source ↗

For the following exercises, use a calculator to solve the equation. Unless indicated otherwise, round all answers to the nearest ten-thousandth.

Newton’s Law of Cooling states that the temperature $T$ of an object at any time t can be described by the equation $T =$ $T_{s} +$ $(T_{0} - T_{s}) e^{- k t},$ where $T_{s}$ is the temperature of the surrounding environment, $T_{0}$ is the initial temperature of the object, and $k$ is the cooling rate. Use the definition of a logarithm along with properties of logarithms to solve the formula for time $t$ such that $t$ is equal to a single logarithm.

Answer

t =

\ln ({(\frac{T - T_{s}}{T_{0} - T_{s}})}^{- \frac{1}{k}})

Try It #1 §6.7 source ↗

The half-life of plutonium-244 is 80,000,000 years. Find a function that gives the amount of plutonium-244 remaining as a function of time, measured in years.

Answer

f (t) =

A_{0} e^{- 0.0000000087 t}

Try It #2 §6.7 source ↗

Cesium-137 has a half-life of about 30 years. If we begin with 200 mg of cesium-137, will it take more or less than 230 years until only 1 milligram remains?

Answer

less than 230 years, 229.3157 to be exact

Try It #3 §6.7 source ↗

Recent data suggests that, as of 2013, the rate of growth predicted by Moore’s Law no longer holds. Growth has slowed to a doubling time of approximately three years. Find the new function that takes that longer doubling time into account.

Answer

f (t) =

A_{0} e^{\frac{\ln 2}{3} t}

Try It #4 §6.7 source ↗

A pitcher of water at 40 degrees Fahrenheit is placed into a 70 degree room. One hour later, the temperature has risen to 45 degrees. How long will it take for the temperature to rise to 60 degrees?

Answer

6.026 hours

Try It #5 §6.7 source ↗

Using the model in Example 6, estimate the number of cases of flu on day 15.

Answer

895 cases on day 15

Try It #6 §6.7 source ↗

Does a linear, exponential, or logarithmic model best fit the data in Table 2? Find the model.

$x$	1	2	3	4	5	6	7	8	9
$y$	3.297	5.437	8.963	14.778	24.365	40.172	66.231	109.196	180.034

Table 2

Answer

Exponential.

y = 2 e^{0.5 x} .

Try It #7 §6.7 source ↗

Change the function

y = 3 {(0.5)}^{x}

to one having

e

as the base.

Answer

y =

3 e^{(\ln 0.5) x}

6.7-1 §6.7 source ↗

With what kind of exponential model would half-life be associated? What role does half-life play in these models?

Answer

Half-life is a measure of decay and is thus associated with exponential decay models. The half-life of a substance or quantity is the amount of time it takes for half of the initial amount of that substance or quantity to decay.

6.7-3 §6.7 source ↗

With what kind of exponential model would doubling time be associated? What role does doubling time play in these models?

Answer

Doubling time is a measure of growth and is thus associated with exponential growth models. The doubling time of a substance or quantity is the amount of time it takes for the initial amount of that substance or quantity to double in size.

6.7-5 §6.7 source ↗

What is an order of magnitude? Why are orders of magnitude useful? Give an example to explain.

Answer

An order of magnitude is the nearest power of ten by which a quantity exponentially grows. It is also an approximate position on a logarithmic scale; Sample response: Orders of magnitude are useful when making comparisons between numbers that differ by a great amount. For example, the mass of Saturn is 95 times greater than the mass of Earth. This is the same as saying that the mass of Saturn is about

10^{2}

times, or 2 orders of magnitude greater, than the mass of Earth.

6.7-7 §6.7 source ↗

For the following exercises, use the logistic growth model $f (x) =$ $\frac{150}{1 + 8 e^{- 2 x}} .$

Find and interpret $f (0) .$ Round to the nearest tenth.

Answer

f (0) \approx 16.7;

The amount initially present is about 16.7 units.

6.7-9 §6.7 source ↗

For the following exercises, use the logistic growth model $f (x) =$ $\frac{150}{1 + 8 e^{- 2 x}} .$

Find the carrying capacity.

Answer

150

6.7-11 §6.7 source ↗

For the following exercises, use the logistic growth model $f (x) =$ $\frac{150}{1 + 8 e^{- 2 x}} .$

Determine whether the data from the table could best be represented as a function that is linear, exponential, or logarithmic. Then write a formula for a model that represents the data.

$x$	$f (x)$
–2	0.694
–1	0.833
0	1
1	1.2
2	1.44
3	1.728
4	2.074
5	2.488

Answer

exponential;

f (x) = {1.2}^{x}

6.7-13 §6.7 source ↗

For the following exercises, enter the data from each table into a graphing calculator and graph the resulting scatter plots. Determine whether the data from the table could represent a function that is linear, exponential, or logarithmic.

$x$	$f (x)$
1	2
2	4.079
3	5.296
4	6.159
5	6.828
6	7.375
7	7.838
8	8.238
9	8.592
10	8.908

Answer

logarithmic

6.7-15 §6.7 source ↗

For the following exercises, enter the data from each table into a graphing calculator and graph the resulting scatter plots. Determine whether the data from the table could represent a function that is linear, exponential, or logarithmic.

$x$	$f (x)$
4	9.429
5	9.972
6	10.415
7	10.79
8	11.115
9	11.401
10	11.657
11	11.889
12	12.101
13	12.295

Answer

logarithmic

6.7-17 §6.7 source ↗

For the following exercises, use a graphing calculator and this scenario: the population of a fish farm in $t$ years is modeled by the equation $P (t) =$ $\frac{1000}{1 + 9 e^{- 0.6 t}} .$

Graph the function.

Answer

6.7-19 §6.7 source ↗

For the following exercises, use a graphing calculator and this scenario: the population of a fish farm in $t$ years is modeled by the equation $P (t) =$ $\frac{1000}{1 + 9 e^{- 0.6 t}} .$

To the nearest tenth, what is the doubling time for the fish population?

Answer

about

1.4

years

6.7-21 §6.7 source ↗

For the following exercises, use a graphing calculator and this scenario: the population of a fish farm in $t$ years is modeled by the equation $P (t) =$ $\frac{1000}{1 + 9 e^{- 0.6 t}} .$

To the nearest tenth, how long will it take for the population to reach $900 ?$

Answer

about

7.3

years

6.7-23 §6.7 source ↗

For the following exercises, use a graphing calculator and this scenario: the population of a fish farm in $t$ years is modeled by the equation $P (t) =$ $\frac{1000}{1 + 9 e^{- 0.6 t}} .$

A substance has a half-life of 2.045 minutes. If the initial amount of the substance was 132.8 grams, how many half-lives will have passed before the substance decays to 8.3 grams? What is the total time of decay?

Answer

4

half-lives;

8.18

minutes

6.7-25 §6.7 source ↗

For the following exercises, use a graphing calculator and this scenario: the population of a fish farm in $t$ years is modeled by the equation $P (t) =$ $\frac{1000}{1 + 9 e^{- 0.6 t}} .$

Recall the formula for calculating the magnitude of an earthquake, $M =$ $\frac{2}{3} \log (\frac{S}{S_{0}}) .$ Show each step for solving this equation algebraically for the seismic moment $S .$

Answer

\begin{array}{l} M = \frac{2}{3} \log (\frac{S}{S_{0}}) \\ \log (\frac{S}{S_{0}}) = \frac{3}{2} M \\ \frac{S}{S_{0}} = 10^{\frac{3 M}{2}} \\ S = S_{0} 10^{\frac{3 M}{2}} \end{array}

6.7-27 §6.7 source ↗

For the following exercises, use a graphing calculator and this scenario: the population of a fish farm in $t$ years is modeled by the equation $P (t) =$ $\frac{1000}{1 + 9 e^{- 0.6 t}} .$

Prove that $b^{x} = e^{x \ln (b)}$ for positive $b \neq 1.$

Answer

Let $y = b^{x}$ for some non-negative real number $b$ such that $b \neq 1.$ Then,

$\begin{array}{l} \ln (y) = \ln (b^{x}) \\ \ln (y) = x \ln (b) \\ e^{\ln (y)} = e^{x \ln (b)} \\ y = e^{x \ln (b)} \end{array}$

6.7-29 §6.7 source ↗

For the following exercises, use this scenario: A doctor prescribes 125 milligrams of a therapeutic drug that decays by about 30% each hour.

Write an exponential model representing the amount of the drug remaining in the patient’s system after $t$ hours. Then use the formula to find the amount of the drug that would remain in the patient’s system after 3 hours. Round to the nearest milligram.

Answer

A =

125 e^{(- 0.3567 t)};

A \approx

43

mg

6.7-31 §6.7 source ↗

For the following exercises, use this scenario: A tumor is injected with $0.5$ grams of Iodine-125, which has a decay rate of $1.15%$ per day.

To the nearest day, how long will it take for half of the Iodine-125 to decay?

Answer

about

60

days

6.7-33 §6.7 source ↗

For the following exercises, use this scenario: A tumor is injected with $0.5$ grams of Iodine-125, which has a decay rate of $1.15%$ per day.

A scientist begins with $250$ grams of a radioactive substance. After $250$ minutes, the sample has decayed to $32$ grams. Rounding to five decimal places, write an exponential equation representing this situation. To the nearest minute, what is the half-life of this substance?

Answer

A (t) =

250 e^{(- 0.00822 t)};

half-life: about

84

minutes

6.7-35 §6.7 source ↗

For the following exercises, use this scenario: A tumor is injected with $0.5$ grams of Iodine-125, which has a decay rate of $1.15%$ per day.

The half-life of Erbium-165 is $10 .4$ hours. What is the hourly decay rate? Express the decimal result to four decimal places and the percentage to two decimal places.

Answer

r \approx - 0.0667,

So the hourly decay rate is about

6.67 %

6.7-37 §6.7 source ↗

For the following exercises, use this scenario: A tumor is injected with $0.5$ grams of Iodine-125, which has a decay rate of $1.15%$ per day.

A research student is working with a culture of bacteria that doubles in size every twenty minutes. The initial population count was $1350$ bacteria. Rounding to five decimal places, write an exponential equation representing this situation. To the nearest whole number, what is the population size after $3$ hours?

Answer

f (t) =

1350 e^{(0.03466 t)};

after 3 hours:

P (180) \approx

691,

200

6.7-39 §6.7 source ↗

For the following exercises, use this scenario: A biologist recorded a count of $360$ bacteria present in a culture after 5 minutes and 1000 bacteria present after 20 minutes.

Rounding to six decimal places, write an exponential equation representing this situation. To the nearest minute, how long did it take the population to double?

Answer

f (t) =

256 e^{(0.068110 t)};

doubling time: about

10

minutes

6.7-41 §6.7 source ↗

For the following exercises, use this scenario: A pot of warm soup with an internal temperature of $100°$ Fahrenheit was taken off the stove to cool in a $69° F$ room. After fifteen minutes, the internal temperature of the soup was $95° F .$

To the nearest minute, how long will it take the soup to cool to $80° F?$

Answer

about

88

minutes

6.7-43 §6.7 source ↗

For the following exercises, use this scenario: A turkey is taken out of the oven with an internal temperature of $165°F$ and is allowed to cool in a $75°F$ room. After half an hour, the internal temperature of the turkey is $145°F .$

Write a formula that models this situation.

Answer

T (t) =

90 e^{(- 0.008377 t)} +

75,

where

t

is in minutes.

6.7-45 §6.7 source ↗

For the following exercises, use this scenario: A turkey is taken out of the oven with an internal temperature of $165°F$ and is allowed to cool in a $75°F$ room. After half an hour, the internal temperature of the turkey is $145°F .$

To the nearest minute, how long will it take the turkey to cool to $110° F?$

Answer

about

113

minutes

6.7-47 §6.7 source ↗

For the following exercises, find the value of the number shown on each logarithmic scale. Round all answers to the nearest thousandth.

Answer

\log (x) =

1.5;

x \approx

31.623

6.7-49 §6.7 source ↗

For the following exercises, find the value of the number shown on each logarithmic scale. Round all answers to the nearest thousandth.

Recall the formula for calculating the magnitude of an earthquake, $M =$ $\frac{2}{3} \log (\frac{S}{S_{0}}) .$ One earthquake has magnitude $3.9$ on the MMS scale. If a second earthquake has $750$ times as much energy as the first, find the magnitude of the second quake. Round to the nearest hundredth.

Answer

MMS magnitude:

5.82

6.7-51 §6.7 source ↗

For the following exercises, use this scenario: The equation $N (t) =$ $\frac{500}{1 + 49 e^{- 0.7 t}}$ models the number of people in a town who have heard a rumor after t days.

To the nearest whole number, how many people will have heard the rumor after 3 days?

Answer

N (3) \approx 71

6.7-53 §6.7 source ↗

For the following exercise, choose the correct answer choice.

A doctor injects a patient with 13 milligrams of radioactive dye that decays exponentially. After 12 minutes, there are 4.75 milligrams of dye remaining in the patient’s system. Which is an appropriate model for this situation?

ⓐ $f (t) =$ $13 {(0.0805)}^{t}$

ⓑ $f (t) =$ $13 e^{0.9195 t}$

ⓓ $f (t) =$ $\frac{4.75}{1 + 13 e^{- 0.83925 t}}$

Answer

C

Try It #1 §6.8 source ↗

Table 2 shows a recent graduate’s credit card balance each month after graduation.

Month	1	2	3	4	5	6	7	8
Debt ($)	620.00	761.88	899.80	1039.93	1270.63	1589.04	1851.31	2154.92

Table 2

ⓐ Use exponential regression to fit a model to these data.
ⓑ If spending continues at this rate, what will the graduate’s credit card debt be one year after graduating?

Answer

ⓐ The exponential regression model that fits these data is $y =$ $522.88585984 {(1.19645256)}^{x} .$

ⓑ If spending continues at this rate, the graduate’s credit card debt will be $4,499.38 after one year.

Try It #2 §6.8 source ↗

Sales of a video game released in the year 2000 took off at first, but then steadily slowed as time moved on. Table 4 shows the number of games sold, in thousands, from the years 2000–2010.

Year	2000	2001	2002	2003	2004	2005
Number Sold (thousands)	142	149	154	155	159	161
Year	2006	2007	2008	2009	2010	-
Number Sold (thousands)	163	164	164	166	167	-

Table 4

ⓐ Let $x$ represent time in years starting with $x = 1$ for the year 2000. Let $y$ represent the number of games sold in thousands. Use logarithmic regression to fit a model to these data.

ⓑ If games continue to sell at this rate, how many games will sell in 2015? Round to the nearest thousand.

Answer

ⓐ The logarithmic regression model that fits these data is $y =$ $141.91242949 +$ $10.45366573 \ln (x)$

ⓑ If sales continue at this rate, about 171,000 games will be sold in the year 2015.

Try It #3 §6.8 source ↗

Table 6 shows the population, in thousands, of harbor seals in the Wadden Sea over the years 1997 to 2012.

Year	Seal Population (Thousands)	Year	Seal Population (Thousands)
1997	3.493	2005	19.590
1998	5.282	2006	21.955
1999	6.357	2007	22.862
2000	9.201	2008	23.869
2001	11.224	2009	24.243
2002	12.964	2010	24.344
2003	16.226	2011	24.919
2004	18.137	2012	25.108

Table 6

ⓐ Let $x$ represent time in years starting with $x = 0$ for the year 1997. Let $y$ represent the number of seals in thousands. Use logistic regression to fit a model to these data.

ⓑ Use the model to predict the seal population for the year 2020.

Answer

ⓐ The logistic regression model that fits these data is $y =$ $\frac{25.65665979}{1 + 6.113686306 e^{- 0.3852149008 x}} .$

ⓑ If the population continues to grow at this rate, there will be about $25,634$ seals in 2020.

6.8-1 §6.8 source ↗

What situations are best modeled by a logistic equation? Give an example, and state a case for why the example is a good fit.

Answer

Logistic models are best used for situations that have limited values. For example, populations cannot grow indefinitely since resources such as food, water, and space are limited, so a logistic model best describes populations.

6.8-3 §6.8 source ↗

What is regression analysis? Describe the process of performing regression analysis on a graphing utility.

Answer

Regression analysis is the process of finding an equation that best fits a given set of data points. To perform a regression analysis on a graphing utility, first list the given points using the STAT then EDIT menu. Next graph the scatter plot using the STAT PLOT feature. The shape of the data points on the scatter graph can help determine which regression feature to use. Once this is determined, select the appropriate regression analysis command from the STAT then CALC menu.

6.8-5 §6.8 source ↗

What does the y-intercept on the graph of a logistic equation correspond to for a population modeled by that equation?

Answer

The y-intercept on the graph of a logistic equation corresponds to the initial population for the population model.

6.8-7 §6.8 source ↗

For the following exercises, match the given function of best fit with the appropriate scatterplot in Figure 7 through Figure 11. Answer using the letter beneath the matching graph.

Figure 7

Figure 8

Figure 9

Figure 10

Figure 11

$y =$ $5.598 -$ $1.912 \ln (x)$

Answer

C

6.8-9 §6.8 source ↗

For the following exercises, match the given function of best fit with the appropriate scatterplot in Figure 7 through Figure 11. Answer using the letter beneath the matching graph.

Figure 7

Figure 8

Figure 9

Figure 10

Figure 11

$y =$ $4.607 +$ $2.733 \ln (x)$

Answer

B

6.8-11 §6.8 source ↗

For the following exercises, match the given function of best fit with the appropriate scatterplot in Figure 7 through Figure 11. Answer using the letter beneath the matching graph.

Figure 7

Figure 8

Figure 9

Figure 10

Figure 11

To the nearest whole number, what is the initial value of a population modeled by the logistic equation $P (t) =$ $\frac{175}{1 + 6.995 e^{- 0.68 t}} ?$ What is the carrying capacity?

Answer

P (0) = 22

; 175

6.8-13 §6.8 source ↗

For the following exercises, match the given function of best fit with the appropriate scatterplot in Figure 7 through Figure 11. Answer using the letter beneath the matching graph.

Figure 7

Figure 8

Figure 9

Figure 10

Figure 11

A logarithmic model is given by the equation $h (p) =$ $67.682 -$ $5.792 \ln (p) .$ To the nearest hundredth, for what value of $p$ does $h (p) = 62 ?$

Answer

p \approx 2.67

6.8-15 §6.8 source ↗

For the following exercises, match the given function of best fit with the appropriate scatterplot in Figure 7 through Figure 11. Answer using the letter beneath the matching graph.

Figure 7

Figure 8

Figure 9

Figure 10

Figure 11

What is the y-intercept on the graph of the logistic model given in the previous exercise?

Answer

y-intercept:

(0, 15)

6.8-17 §6.8 source ↗

For the following exercises, use this scenario: The population $P$ of a koi pond over $x$ months is modeled by the function $P (x) =$ $\frac{68}{1 + 16 e^{- 0.28 x}} .$

What was the initial population of koi?

Answer

4

koi

6.8-19 §6.8 source ↗

For the following exercises, use this scenario: The population $P$ of a koi pond over $x$ months is modeled by the function $P (x) =$ $\frac{68}{1 + 16 e^{- 0.28 x}} .$

How many months will it take before there are $20$ koi in the pond?

Answer

about

6.8

months.

6.8-21 §6.8 source ↗

For the following exercises, use this scenario: The population $P$ of an endangered species habitat for wolves is modeled by the function $P (x) =$ $\frac{558}{1 + 54.8 e^{- 0.462 x}},$ where $x$ is given in years.

Graph the population model to show the population over a span of $10$ years.

Answer

A two-dimensional line graph with an x-axis ranging from 0 to 20 and a y-axis ranging from 0 to 600. A dark blue, S-shaped curve begins near the origin, approximately at (1, 15), and ascends through the graph, passing roughly through points (5, 80), (10, 300), and (15, 500), before ending with an arrow around (17.5, 550) towards the top right.

6.8-23 §6.8 source ↗

For the following exercises, use this scenario: The population $P$ of an endangered species habitat for wolves is modeled by the function $P (x) =$ $\frac{558}{1 + 54.8 e^{- 0.462 x}},$ where $x$ is given in years.

How many wolves will the habitat have after $3$ years?

Answer

About 38 wolves

6.8-25 §6.8 source ↗

For the following exercises, use this scenario: The population $P$ of an endangered species habitat for wolves is modeled by the function $P (x) =$ $\frac{558}{1 + 54.8 e^{- 0.462 x}},$ where $x$ is given in years.

Use the intersect feature to approximate the number of years it will take before the population of the habitat reaches half its carrying capacity.

Answer

About 8.7 years

6.8-27 §6.8 source ↗

For the following exercises, refer to Table 7.

x	1	2	3	4	5	6
*f(x)*	1125	1495	2310	3294	4650	6361

Table 7

Use the regression feature to find an exponential function that best fits the data in the table.

Answer

f (x) =

776.682 {(1.426)}^{x}

6.8-29 §6.8 source ↗

For the following exercises, refer to Table 7.

x	1	2	3	4	5	6
*f(x)*	1125	1495	2310	3294	4650	6361

Table 7

Graph the exponential equation on the scatter diagram.

Answer

A graph illustrating exponential growth, showing an orange curve with six blue data points plotted. As the x-value increases, the y-value rapidly increases, indicating a steep upward trend.

6.8-31 §6.8 source ↗

For the following exercises, refer to Table 8.

x	1	2	3	4	5	6
*f(x)*	555	383	307	210	158	122

Table 8

Use a graphing calculator to create a scatter diagram of the data.

Answer

A scatter plot on a grid shows six blue data points with x-values from 1 to 6 and corresponding y-values decreasing from 550 to 120, illustrating a negative correlation.

6.8-33 §6.8 source ↗

For the following exercises, refer to Table 8.

x	1	2	3	4	5	6
*f(x)*	555	383	307	210	158	122

Table 8

Write the exponential function as an exponential equation with base $e .$

Answer

f (x) =

{731.92 e}^{-0.3038 x}

6.8-35 §6.8 source ↗

For the following exercises, refer to Table 8.

x	1	2	3	4	5	6
*f(x)*	555	383	307	210	158	122

Table 8

Use the intersect feature to find the value of $x$ for which $f (x) = 250.$

Answer

When

f (x) =

250,

x \approx

3.6

6.8-37 §6.8 source ↗

For the following exercises, refer to Table 9.

x	1	2	3	4	5	6
*f(x)*	5.1	6.3	7.3	7.7	8.1	8.6

Table 9

Use the LOGarithm option of the REGression feature to find a logarithmic function of the form $y = a + b \ln (x)$ that best fits the data in the table.

Answer

y =

5.063 +

1.934 log (x)

6.8-39 §6.8 source ↗

For the following exercises, refer to Table 9.

x	1	2	3	4	5	6
*f(x)*	5.1	6.3	7.3	7.7	8.1	8.6

Table 9

Graph the logarithmic equation on the scatter diagram.

Answer

An image showing an orange curve passing through six blue data points on a grid with x-axis from 0 to 7 and y-axis from 0 to 10, indicating an increasing trend.

6.8-41 §6.8 source ↗

For the following exercises, refer to Table 10.

x	1	2	3	4	5	6	7	8
*f(x)*	7.5	6	5.2	4.3	3.9	3.4	3.1	2.9

Table 10

Use a graphing calculator to create a scatter diagram of the data.

Answer

A scatter plot shows a series of data points illustrating a decreasing trend. As the x-value increases from 1 to 8, the corresponding y-value generally decreases from approximately 7.5 to 2.9, indicating a negative correlation.

6.8-43 §6.8 source ↗

For the following exercises, refer to Table 10.

x	1	2	3	4	5	6	7	8
*f(x)*	7.5	6	5.2	4.3	3.9	3.4	3.1	2.9

Table 10

Use the logarithmic function to find the value of the function when $x = 10.$

Answer

When

f (10) \approx 2.3

6.8-45 §6.8 source ↗

For the following exercises, refer to Table 10.

x	1	2	3	4	5	6	7	8
*f(x)*	7.5	6	5.2	4.3	3.9	3.4	3.1	2.9

Table 10

Use the intersect feature to find the value of $x$ for which $f (x) = 8.$

Answer

When

f (x) =

8,

x \approx

0.82

6.8-47 §6.8 source ↗

For the following exercises, refer to Table 11.

x	1	2	3	4	5	6	7	8	9	10
*f(x)*	8.7	12.3	15.4	18.5	20.7	22.5	23.3	24	24.6	24.8

Table 11

Use the LOGISTIC regression option to find a logistic growth model of the form $y = \frac{c}{1 + a e^{- b x}}$ that best fits the data in the table.

Answer

f (x) =

\frac{25.081}{1 + 3.182 e^{- 0.545 x}}

6.8-49 §6.8 source ↗

For the following exercises, refer to Table 11.

x	1	2	3	4	5	6	7	8	9	10
*f(x)*	8.7	12.3	15.4	18.5	20.7	22.5	23.3	24	24.6	24.8

Table 11

To the nearest whole number, what is the predicted carrying capacity of the model?

Answer

About 25

6.8-51 §6.8 source ↗

For the following exercises, refer to Table 12.

$x$	0	2	4	5	7	8	10	11	15	17
$f (x)$	12	28.6	52.8	70.3	99.9	112.5	125.8	127.9	135.1	135.9

Table 12

Use a graphing calculator to create a scatter diagram of the data.

Answer

A scatter plot shows ten data points on a grid with an x-axis from 0 to 18 and a y-axis from 0 to 140. The points exhibit a strong increasing trend.

6.8-53 §6.8 source ↗

For the following exercises, refer to Table 12.

$x$	0	2	4	5	7	8	10	11	15	17
$f (x)$	12	28.6	52.8	70.3	99.9	112.5	125.8	127.9	135.1	135.9

Table 12

Graph the logistic equation on the scatter diagram.

Answer

A graph with an x-axis labeled from 0 to 18 and a y-axis labeled from 0 to 140. An orange S-shaped curve is plotted, starting near (0, 11), increasing steeply through points such as (2, 29), (4, 53), (5, 70), (7, 100), (8, 112), and (10, 126), and then flattening out as it approaches a maximum y-value around 135-136, with data points at (11, 128) and (15, 134). The curve suggests logistic growth.

6.8-55 §6.8 source ↗

For the following exercises, refer to Table 12.

$x$	0	2	4	5	7	8	10	11	15	17
$f (x)$	12	28.6	52.8	70.3	99.9	112.5	125.8	127.9	135.1	135.9

Table 12

Use the intersect feature to find the value of $x$ for which the model reaches half its carrying capacity.

Answer

When

f (x) =

68,

x \approx

4.9

6.8-57 §6.8 source ↗

For the following exercises, refer to Table 12.

$x$	0	2	4	5	7	8	10	11	15	17
$f (x)$	12	28.6	52.8	70.3	99.9	112.5	125.8	127.9	135.1	135.9

Table 12

Use a graphing utility to find an exponential regression formula $f (x)$ and a logarithmic regression formula $g (x)$ for the points $(1.5, 1.5)$ and $(8.5, 8.5) .$ Round all numbers to 6 decimal places. Graph the points and both formulas along with the line $y = x$ on the same axis. Make a conjecture about the relationship of the regression formulas.

Answer

f (x) =

1.034341 {(1.281204)}^{x}

;

g (x) =

4.035510

; the regression curves are symmetrical about

y = x

, so it appears that they are inverse functions.

6.8-59 §6.8 source ↗

For the following exercises, refer to Table 12.

$x$	0	2	4	5	7	8	10	11	15	17
$f (x)$	12	28.6	52.8	70.3	99.9	112.5	125.8	127.9	135.1	135.9

Table 12

Find the inverse function $f^{- 1} (x)$ for the logistic function $f (x) =$ $\frac{c}{1 + a e^{- b x}} .$ Show all steps.

Answer

f^{- 1} (x) =

\frac{ln (a) - ln (\frac{c}{x} - 1)}{b}

6.R-1 §6.R source ↗

Determine whether the function

y =

156 {(0.825)}^{t}

represents exponential growth, exponential decay, or neither. Explain

Answer

exponential decay; The growth factor,

0.825,

is between

0

and

1.

6.R-3 §6.R source ↗

Find an exponential equation that passes through the points

(2, 2 .25)

and

(5, 60.75) .

Answer

y = 0.25 {(3)}^{x}

6.R-5 §6.R source ↗

A retirement account is opened with an initial deposit of $8,500 and earns

8.12 %

interest compounded monthly. What will the account be worth in

20

years?

Answer

$ 42, 888.18

6.R-7 §6.R source ↗

Does the equation

y =

2.294 e^{- 0.654 t}

represent continuous growth, continuous decay, or neither? Explain.

Answer

continuous decay; the growth rate is negative.

6.R-9 §6.R source ↗

Graph the function

f (x) =

3.5 {(2)}^{x} .

State the domain and range and give the y-intercept.

Answer

domain: all real numbers; range: all real numbers strictly greater than zero; y-intercept: (0, 3.5);

6.R-11 §6.R source ↗

The graph of

f (x) = {6.5}^{x}

is reflected about the y-axis and stretched vertically by a factor of

7.

What is the equation of the new function,

g (x) ?

State its y-intercept, domain, and range.

Answer

g (x) =

7 {(6.5)}^{- x};

y-intercept:

(0, 7);

Domain: all real numbers; Range: all real numbers greater than

0.

6.R-13 §6.R source ↗

Rewrite

\log_{17} (4913) =

x

as an equivalent exponential equation.

Answer

17^{x} = 4913

6.R-15 §6.R source ↗

Rewrite

a^{- \frac{2}{5}} = b

as an equivalent logarithmic equation.

Answer

\log_{a} b = - \frac{2}{5}

6.R-17 §6.R source ↗

Solve for x if

\log_{64} (x) = \frac{1}{3}

by converting the logarithmic equation

\log_{64} (x) = \frac{1}{3}

to exponential form.

Answer

x = 64^{\frac{1}{3}} = 4

6.R-19 §6.R source ↗

Evaluate

\log (0.000001)

without using a calculator.

Answer

\log (0 .000001) =

-

6

6.R-21 §6.R source ↗

Evaluate

\ln (e^{- 0.8648})

without using a calculator.

Answer

\ln (e^{- 0.8648}) =

-

0.8648

6.R-23 §6.R source ↗

Graph the function

g (x) =

\log (7 x + 21) -

4.

Answer

6.R-25 §6.R source ↗

State the domain, vertical asymptote, and end behavior of the function

g (x) =

\ln (4 x + 20) -

17.

Answer

Domain:

x > - 5;

Vertical asymptote:

x = - 5;

End behavior: as

x \to

-

5^{+},

f (x) \to

-

\infty

and as

x \to \infty, f (x) \to \infty .

6.R-27 §6.R source ↗

Rewrite

\log_{8} (x) +

\log_{8} (5) +

\log_{8} (y) +

\log_{8} (13)

in compact form.

Answer

{log}_{8} (65 x y)

6.R-29 §6.R source ↗

Rewrite

\ln (z) -

\ln (x) -

\ln (y)

in compact form.

Answer

\ln (\frac{z}{x y})

6.R-31 §6.R source ↗

Rewrite

- \log_{y} (\frac{1}{12})

as a single logarithm.

Answer

{log}_{y} (12)

6.R-33 §6.R source ↗

Use properties of logarithms to expand

\ln (2 b \sqrt{\frac{b + 1}{b - 1}}) .

Answer

\ln (2) +

\ln (b) +

\frac{\ln (b + 1) - \ln (b - 1)}{2}

6.R-35 §6.R source ↗

Condense the expression

3 \log_{7} v +

6 \log_{7} w -

\frac{\log_{7} u}{3}

to a single logarithm.

Answer

\log_{7} (\frac{v^{3} w^{6}}{\sqrt[3]{u}})

6.R-37 §6.R source ↗

Rewrite

5^{12 x - 17} = 125

as a logarithm. Then apply the change of base formula to solve for

x

using the common log. Round to the nearest thousandth.

Answer

x =

\frac{\frac{\log (125)}{\log (5)} + 17}{12} =

\frac{5}{3}

6.R-39 §6.R source ↗

Solve

\frac{125}{{(\frac{1}{625})}^{- x - 3}} =

5^{3}

by rewriting each side with a common base.

Answer

x = - 3

6.R-41 §6.R source ↗

Use logarithms to find the exact solution for

3 e^{6 n - 2} +

1 =

-

60.

If there is no solution, write no solution.

Answer

no solution

6.R-43 §6.R source ↗

Find the exact solution for

2 e^{5 x - 2} -

9 =

-

56.

If there is no solution, write no solution.

Answer

no solution

6.R-45 §6.R source ↗

Find the exact solution for

e^{2 x} -

e^{x} -

110 =

0.

If there is no solution, write no solution.

Answer

x = \ln (11)

6.R-47 §6.R source ↗

Use the definition of a logarithm to find the exact solution for

9 +

6 \ln (a + 3) =

33.

Answer

a = e^{4} - 3

6.R-49 §6.R source ↗

Use the one-to-one property of logarithms to find an exact solution for

\ln (5) +

\ln (5 x^{2} - 5) =

\ln (56) .

If there is no solution, write no solution.

Answer

x = \pm \frac{9}{5}

6.R-51 §6.R source ↗

The population of a city is modeled by the equation

P (t) =

256,

114 e^{0.25 t}

where

t

is measured in years. If the city continues to grow at this rate, how many years will it take for the population to reach one million?

Answer

about

5.45

years

6.R-53 §6.R source ↗

Find the inverse function

f^{- 1}

for the logarithmic function

f (x) =

0.25 \cdot \log_{2} (x^{3} + 1) .

Answer

f^{- 1} (x) =

\sqrt[3]{2^{4 x} - 1}

6.R-55 §6.R source ↗

For the following exercises, use this scenario: A doctor prescribes $300$ milligrams of a therapeutic drug that decays by about $17 %$ each hour.

Write an exponential model representing the amount of the drug remaining in the patient’s system after $t$ hours. Then use the formula to find the amount of the drug that would remain in the patient’s system after $24$ hours. Round to the nearest hundredth of a gram.

Answer

f (t) =

300 {(0.83)}^{t};

f (24) \approx 3.43 g

6.R-57 §6.R source ↗

For the following exercises, use this scenario: A soup with an internal temperature of $350°$ Fahrenheit was taken off the stove to cool in a $71°F$ room. After fifteen minutes, the internal temperature of the soup was $175°F .$

How many minutes will it take the soup to cool to $85°F?$

Answer

about

45

minutes

6.R-59 §6.R source ↗

For the following exercises, use this scenario: The equation $N (t) =$ $\frac{1200}{1 + 199 e^{- 0.625 t}}$ models the number of people in a school who have heard a rumor after $t$ days.

To the nearest tenth, how many days will it be before the rumor spreads to half the carrying capacity?

Answer

about

8.5

days

6.R-61 §6.R source ↗

For the following exercises, enter the data from each table into a graphing calculator and graph the resulting scatter plots. Determine whether the data from the table would likely represent a function that is linear, exponential, or logarithmic.

x	*f(x)*
1	3.05
2	4.42
3	6.4
4	9.28
5	13.46
6	19.52
7	28.3
8	41.04
9	59.5
10	86.28

Answer

exponential

6.R-63 §6.R source ↗

For the following exercises, enter the data from each table into a graphing calculator and graph the resulting scatter plots. Determine whether the data from the table would likely represent a function that is linear, exponential, or logarithmic.

Find a formula for an exponential equation that goes through the points $(- 2, 100)$ and $(0, 4) .$ Then express the formula as an equivalent equation with base e.

Answer

y = 4 {(0.2)}^{x};

y =

4 e^{-1.609438 x}

6.R-65 §6.R source ↗

For the following exercises, enter the data from each table into a graphing calculator and graph the resulting scatter plots. Determine whether the data from the table would likely represent a function that is linear, exponential, or logarithmic.

The population of a culture of bacteria is modeled by the logistic equation $P (t) =$ $\frac{14, 250}{1 + 29 e^{- 0.62 t}},$ where $t$ is in days. To the nearest tenth, how many days will it take the culture to reach $75 %$ of its carrying capacity?

Answer

about

7.2

days

6.R-67 §6.R source ↗

For the following exercises, use a graphing utility to create a scatter diagram of the data given in the table. Observe the shape of the scatter diagram to determine whether the data is best described by an exponential, logarithmic, or logistic model. Then use the appropriate regression feature to find an equation that models the data. When necessary, round values to five decimal places.

x	*f(x)*
0.15	36.21
0.25	28.88
0.5	24.39
0.75	18.28
1	16.5
1.5	12.99
2	9.91
2.25	8.57
2.75	7.23
3	5.99
3.5	4.81

Answer

logarithmic; $y =$ $16.68718 -$ $9.71860 \ln (x)$

6.T-1 §6.T source ↗

The population of a pod of bottlenose dolphins is modeled by the function

A (t) =

8 {(1.17)}^{t},

where

t

is given in years. To the nearest whole number, what will the pod population be after

3

years?

Answer

About 13 dolphins.

6.T-3 §6.T source ↗

Drew wants to save $2,500 to go to the next World Cup. To the nearest dollar, how much will he need to invest in an account now with

6.25 %

APR, compounding daily, in order to reach his goal in

4

years?

Answer

$ 1,947

6.T-5 §6.T source ↗

Graph the function

f (x) =

5 {(0.5)}^{- x}

and its reflection across the y-axis on the same axes, and give the y-intercept.

Answer

y-intercept: $(0, 5)$

Graph of f(-x)=5(0.5)^-x in blue and f(x)=5(0.5)^x in orange.

6.T-7 §6.T source ↗

Rewrite

\log_{8.5} (614.125) =

a

as an equivalent exponential equation.

Answer

{8.5}^{a} =

614.125

6.T-9 §6.T source ↗

Solve for

x

by converting the logarithmic equation

l o g_{\frac{1}{7}} (x) = 2

to exponential form.

Answer

x = {(\frac{1}{7})}^{2} = \frac{1}{49}

6.T-11 §6.T source ↗

Evaluate

\ln (0.716)

using a calculator. Round to the nearest thousandth.

Answer

\ln (0.716) \approx

-

0.334

6.T-13 §6.T source ↗

State the domain, vertical asymptote, and end behavior of the function

f (x) =

\log_{5} (39 - 13 x) +

7.

Answer

Domain:

x < 3;

Vertical asymptote:

x = 3;

End behavior:

x \to

3^{-},

f (x) \to

-

\infty

and

x \to - \infty, f (x) \to \infty

6.T-15 §6.T source ↗

Rewrite

\log_{t} (96) -

\log_{t} (8)

in compact form.

Answer

\log_{t} (12)

6.T-17 §6.T source ↗

Use properties of logarithm to expand

\ln (y^{3} z^{2} \cdot \sqrt[3]{x - 4}) .

Answer

3 \ln (y) +

2 \ln (z) +

\frac{\ln (x - 4)}{3}

6.T-19 §6.T source ↗

Rewrite

16^{3 x - 5} = 1000

as a logarithm. Then apply the change of base formula to solve for

x

using the natural log. Round to the nearest thousandth.

Answer

x =

\frac{\frac{\ln (1000)}{\ln (16)} + 5}{3} \approx

2.497

6.T-21 §6.T source ↗

Use logarithms to find the exact solution for

-

9 e^{10 a - 8} -

5 =

-

41

. If there is no solution, write no solution.

Answer

a = \frac{\ln (4) + 8}{10}

6.T-23 §6.T source ↗

Find the exact solution for

-

5 e^{- 4 x - 1} -

4 =

64.

If there is no solution, write no solution.

Answer

no solution

6.T-25 §6.T source ↗

Find the exact solution for

e^{2 x} -

e^{x} -

72 =

0.

If there is no solution, write no solution.

Answer

x = \ln (9)

6.T-27 §6.T source ↗

Use the one-to-one property of logarithms to find an exact solution for

\log (4 x^{2} - 10) +

\log (3) =

\log (51)

If there is no solution, write no solution.

Answer

x = \pm \frac{3 \sqrt{3}}{2}

6.T-29 §6.T source ↗

A radiation safety officer is working with

112

grams of a radioactive substance. After

17

days, the sample has decayed to

80

grams. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest day, what is the half-life of this substance?

Answer

f (t) =

112 e^{- .019792 t};

half-life: about

35

days

6.T-31 §6.T source ↗

A bottle of soda with a temperature of

71°

Fahrenheit was taken off a shelf and placed in a refrigerator with an internal temperature of

35° F .

After ten minutes, the internal temperature of the soda was

63° F .

Use Newton’s Law of Cooling to write a formula that models this situation. To the nearest degree, what will the temperature of the soda be after one hour?

Answer

T (t) =

36 e^{- 0.025131 t} +

35;

T (60) \approx

43^{o} F

6.T-33 §6.T source ↗

Enter the data from Table 1 into a graphing calculator and graph the resulting scatter plot. Determine whether the data from the table would likely represent a function that is linear, exponential, or logarithmic.

x	*f(x)*
1	3
2	8.55
3	11.79
4	14.09
5	15.88
6	17.33
7	18.57
8	19.64
9	20.58
10	21.42

Table 1

Answer

logarithmic

6.T-35 §6.T source ↗

For the following exercises, use a graphing utility to create a scatter diagram of the data given in the table. Observe the shape of the scatter diagram to determine whether the data is best described by an exponential, logarithmic, or logistic model. Then use the appropriate regression feature to find an equation that models the data. When necessary, round values to five decimal places.

x	*f(x)*
1	20
2	21.6
3	29.2
4	36.4
5	46.6
6	55.7
7	72.6
8	87.1
9	107.2
10	138.1

Answer

exponential; $y =$ $15.10062 {(1.24621)}^{x}$

6.T-37 §6.T source ↗

For the following exercises, use a graphing utility to create a scatter diagram of the data given in the table. Observe the shape of the scatter diagram to determine whether the data is best described by an exponential, logarithmic, or logistic model. Then use the appropriate regression feature to find an equation that models the data. When necessary, round values to five decimal places.

x	*f(x)*
0	2.2
0.5	2.9
1	3.9
1.5	4.8
2	6.4
3	9.3
4	12.3
5	15
6	16.2
7	17.3
8	17.9

Answer

logistic; $y =$ $\frac{18.41659}{1 + 7.54644 e^{- 0.68375 x}}$