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Exponential and Logarithmic Functions

CHAPTER

12.1 Composite and Inverse Functions12.2 Exponential Functions12.3 Logarithmic Functions12.4 Properties of Logarithms12.5 Common and Natural Logarithms12.6 Exponential and Logarithmic

Equations with Applications

Composite and Inverse Functions12.112.1

1. Find the compososition of two functions.2. Show that two functions are inverses.3. Show that a function is one-to-one.4. Find the inverse of a function. 5. Graph a given function’s inverse function.

Composite functions: If f and g are functions, then the composition of f and g is defined asFor all x in the domain g for which g(x) is the domain of f. The composition of g and f is defined as

for all x in the domain of g for which f(x) is in the domain of g.

f g x f g x

g f x g f x

If and find the following.

Solution

Example

( ) 3 8f x x ( ) 2 5,g x x

f g x f g x

2 5f x

53 2 8x

6 15 8x

Replace g(x) with 2x – 5.

In f(x), replace x with 2x – 5.

Simplify.

a. b. g f x

continued

g f x g f x

3 8g x

6 16 5x

Replace f(x) with 3x + 8.

In g(x), replace x with 3x + 8

Simplify.

82 3 5x

Inverse functions: Two functions f and g are inverses if and only if for all x in the domain of g and for all x in the domain of f.

f g x x g f x x

Inverse FunctionsTo determine whether two functions f and g are inverses of each other,

1. Show that for all x in the domain of g. 2. Show that for all x in the domain of f.

f g x x g f x x

Solution

Example

Verify that f and g are inverses.

3( ) 4 3, ( )

xf x x g x

We need to show that and f g x x .g f x x 3

g f x 4 3g x

4 3 3x

Since and , f and g are inverses. f g x x g f x x

One-to-One function: A function f is one-to-one if for any two numbers a and b in its domain, when f(a) = f(b), a = b and when a b, f(a) f(b).

Horizontal Line Test for One-to-One FunctionsGiven a function’s graph, the function is one-to-one if every horizontal line that can intersect the graph does so at one and only one point.

Determine whether the graph is a one-to-one function.

Solution

Example

2 5f x x

-5 -4 -3 -2 -1 1 2 3 4 5

A horizontal line can intersect the graph in more than one point, so the function is not one-to-one.

Finding the Inverse Function of a One-to-One Function

1. If necessary, replace f(x) with y. 2. Replace all x’s with y’s and y’s with x’s.3. Solve the equation from step 2 for y.4. Replace y with f-1(x).

Existence of Inverse FunctionsA function has an inverse function if and only if the function is one-to-one.

Solution

Example

Find f -1(x) for the function f(x) = 7x – 4.

y = 7x – 4

x = 7y – 44

Since f[f -1(x)]= x and f -1[f (x)] = x, they are inverses.

Graph of Inverse FunctionsThe graphs of f and f-1 are symmetric to the graph of y = x.

Solution

Example

Sketch the inverse of the function whose graph is shown.

Draw the line y = x and reflect the graph on the line.

-10 -8 -6 -4 -2 2 4 6 8 10

If f(x) = x + 7 and g(x) = 2x – 12, what is

a) -44

4 .f g

If f(x) = x + 7 and g(x) = 2x – 12, what is

a) -44

4 .f g

Find f -1(x) for 6x – 7.

Exponential Functions12.212.2

1. Define and graph exponential functions.2. Solve equations of the form bx = by for x.3. Use exponential functions to solve application problems.

Exponential function: If b > 0, b 1, and x is any real number, then the exponential function is f(x) = bx.

Note: The definition of the exponential function has two restrictions on b. If b = 1, then f(x) = bx = 1x = 1, which is a linear function. If b < 0, then we could get values for which the function is not defined as a real number.

ExampleGraph f(x) = 2x, and g(x) = 3x.. Solution

x –2 –1 0 1 2

f(x) 1 2 4

g(x) 1 3 9

Comparing the graphs, we can see the greater value of b, the steeper the graph.

f(x) = 2xg(x) = 3x

-10-9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10

-10-9-8-7-6-5-4-3-2-1

123456789

ExampleGraph f(x) = 3x+3 .

Solutionx f(x)

-4 3-1 =

-3 30 = 1

-2 31 = 3

-1 32 = 9

0 33 = 27

f(x) = 3x+3

-24 -21 -18 -15 -12 -9 -6 -3 3

Solving Exponential Equations1. If necessary, write both sides of the equation as a

power of the same base.2. If necessary, simplify the exponents.3. Set the exponents equal to each other.4. Solve the resulting equation.

The One-to-One Property of ExponentialsGiven b > 0 and b 1, if bx = by, then x = y.

Example

Solve.

a. 3x = 81 b. 2 14

Solution

a. 3x = 81

3x = 34

The solution set is {4}.

b. 2 14

2 24 4x

The solution set is {–4}.

Example

Rayanne deposited $15,000 in an account that pays 7% annual interest compounded quarterly. How much is accumulated in the account after 10 years?

SolutionUnderstand We are asked to find A and given that

t = 10, P = $15,000, r = 0.07, and n = 4.

Plan Use the formula 1 .nt

continued

4(10)0.07

5 11 004

4015,000 1 0.0175A

$30,023.96A

Substitution.

Simplify.

Evaluate using a calculator and round to the nearest cent.

Execute

Answer After 10 years, the accumulated account is $30,023.96.

continued

4(10)0.07

30,023.96 14

30,023.96 1.0175P

14,999.99993 P

Check Verify that the principal is $15,000 if the accumulated amount is $30,023.96 after the principal is compounded quarterly.

30,023.96

1.0175P

Since the accumulated amount was rounded, it is expected our calculated value of the principal to be slightly different from $15,000.

Which of the following functions does the graph represent?

a) f(x) = 4x+2

b) f(x) = 4x–2

c) f(x) = 2x+4

d) f(x) = 2x–4

Which of the following functions does the graph represent?

a) f(x) = 4x+2

b) f(x) = 4x–2

c) f(x) = 2x+4

d) f(x) = 2x–4

A stereo system is purchased for $2500. It’s value each year is about 85% of its value in the preceding year. Its value in dollars after t years is given by the exponential function V(t) = 2500(0.85)t. Find the salvage value of the stereo after 4 yr.

a) $1285.23

b) $1296.42

c) $1305.02

d) $1316.84

A stereo system is purchased for $2500. It’s value each year is about 85% of its value in the preceding year. Its value in dollars after t years is given by the exponential function V(t) = 2500(0.85)t. Find the salvage value of the stereo after 4 yr.

a) $1285.23

b) $1296.42

c) $1305.02

d) $1316.84

Logarithmic Functions12.312.3

1. Convert between exponential and logarithmic forms.2. Solve logarithmic equations by changing to

exponential form.3. Graph logarithmic functions.4. Solve applications involving logarithms.

Logarithm: If b > 0 and b 1, then y = logbx is equivalent to x = by.

x = by y = logbx

The exponent is the logarithm.

The base is the base of the logarithm.

Solving Logarithmic EquationsTo solve an equation of the form logbx = y, where b, x, or y is a variable, write the equation in exponential form, by = x, and then solve for the variable.

Example

Solve log7 x = 3.

SolutionWrite the equation in exponential form. log7 x = 3

73 = x 343 = x

ExampleSolve

Solution Begin by rewriting the equation in exponential form and then solve.

Divide both sides by -3 to isolate the logarithm.

Simplify.

Subtract 11 from both sides.

211 3log 4x

23log 15x

2log 5x 52x

Write in exponential form.

For any real number b, where b > 0 and b 1, 1. logbb = 1 2. log b 1 = 0

Graphing Logarithmic FunctionsTo graph a function in the form f(x) = logbx,

1. Replace f(x) with y and then write the logarithm in exponential form x = by.

2. Find ordered pairs that satisfy the equation by assigning values to y and finding x.

3. Plot the ordered pairs and draw a smooth curve through the points.

ExampleGraph f(x) = log1/4x.

SolutionReplacing f(x) by y, we see that the equation y = log1/4 x can be rewritten in exponential form as

214 16

11 14 4

21 14 16

ExampleThe function P = 95 – 30 log2x models the percent, P, of students who recall the important features of a classroom lecture over time, where x is the number of days that have elapsed since the lecture was given. What percent of the students recall the important features of a lecture 8 days after it was given? (Source: Psychology for the New Millennium, 8th Edition, Spencer A. Rathos, Thomson Publishing Company)

SolutionUnderstand We are given the function that models the

percent, P, of students who recall the important features of a lecture x days after it is given. We are to find the percent of students who recall the important features of a lecture 8 days after it was given.

continued

Execute

Answer 5% of the students remember the important features of a lecture 8 days after it is given.

Plan Evaluate P = 95 – 30log2 x where x = 8.

P = 95 – 30log2 8

P = 95 – 30(3)

P = 95 – 90

Check 5 = 95 – 30log2 x

–90 = – 30log2 x

3 = log2 x

23 = x

Write 62 = 36 in logarithmic form.

6log 2 36

6log 36 2

2log 6 36

2log 36 6

Write 62 = 36 in logarithmic form.

6log 2 36

6log 36 2

2log 6 36

2log 36 6

1log 4

Solve log3x = 5.

a) -243

b) -125

c) 125

d) 243

Solve log3x = 5.

a) -243

b) -125

c) 125

d) 243

Properties of Logarithms12.412.4

1. Apply the inverse properties of logarithms.2. Apply the product, quotient, and power properties of

logarithms.

Inverse Properties of LogarithmsFor any real numbers b and x, where b > 0 and b 1,

and x > 0,1. blog

bx = x 2. logbbx= x

Further Properties of LogarithmsFor real numbers x, y, and b, where x > 0, y > 0, b > 0,

and b 1.Product Rule of Logarithms: logbxy = logbx + logby

Quotient Rule of Logarithms:

Power Rule of Logarithms: logbxr = rlogbx

(The logarithm of the product of two numbers is equal to the sum of the logarithms of the numbers.)

log log logb b b

(The logarithm of the quotient of two numbers is equal to the difference of the logarithms of the numbers.)

(The logarithm of a number raised to the power is equal to the exponent times the logarithm of the number.)

Example

Use the product rule of logarithms to write each expression as a sum of logarithms.

a. logbpqr b. logb x(x + 5)

Solution

a. logbpqr = logbp + logbq + logbr

b. logb x(x+5) = logbx + logb(x + 5)

Example

Use the quotient rule of logarithms in the form to write the expression as a

single logarithm.

Solution

log log logb b b

6 6log 108 log x

6 6log 108 log x 6

108log

Example

Use the power rule of logarithms to write the expression as a multiple of a logarithm.

Solution

a. log3x11

log3x11 = 11 log3x

1logb x

= logbx-7 = -7logbx

Example

Write the expression as a sum or difference of multiples of logarithms.

Solution

Use the product and quotient rules.

= logb x7+ logb z9 logb y4

Use the power rule.

= 7 logb x + 9 logb z 4 logb y

Example

Write the expression as a single logarithm. Leave answer in simplest form without negative or fractional exponents.

Solution log 5 log 4a ax x

log 5 log 4a ax x log 5 4a x x

2log 20a x x

Use the power rule to write the expression as a multiple of a logarithm.

log 8a

1log 2

1log 8

Use the power rule to write the expression as a multiple of a logarithm.

log 8a

1log 2

1log 8

Express as a single logarithm:

log 2log 5log .a a ax y z

2 5loga

log 2 5a x y z

Express as a single logarithm:

log 2log 5log .a a ax y z

2 5loga

log 2 5a x y z

Common and Natural Logarithms12.512.5

1. Define common logarithms and evaluate them using a calculator.

2. Solve applications using common logarithms.3. Define natural logarithms and evaluate them using a

calculator.4. Solve applications using natural logarithms.

Common logarithms: Logarithms with a base of 10. Log10x is written as log x.

Solution

Example

Use a calculator to approximate each common logarithm. Round to the nearest thousandth if necessary. a. log 456 b. log 0.00257

a. log 456

log 0.00257 2.590

ExampleThe function is used to calculate sound

intensity, where d represents the intensity in decibels, I represents the intensity watts per unit of area, and I0 represents the faintest audible sound to the average human ear, which is 10-12 watts per square meter. What is the intensity level of sounds at a decibel level of 75 dB?

010log

Understand We are given the function, and we are to find the sound intensity, I. 0

10log ,I

continued

Execute

Answer The sound intensity is 10–4.5.

Plan Using , substitute 75 for d and

10-12 for I0 and then solve for I.0

10logI

1275 10 log

127.5 log

12 12 7.512

10 10 1010

4.510 I

continued

Check If the sound intensity is 10-4.5, verify the decibel reading is 75.

10log10

7.510log10d

10 7.5d

Natural logarithms: Base-e logarithms are called natural logarithms and logex is written as ln x. Note that ln e = 1.

Solution

ExampleUse a calculator to approximate each natural logarithm to four decimal places. a. ln 67 b. ln 0.0072

a. ln 67

b. ln 0.00257 5.9638

4.2047

ExampleIf an account pays 8% annual interest, compounded continuously, how long will it take a deposit of $25,000 to produce an account balance of $100,000?Understand We are to find the time it takes for $25,000

to grow to $100,000 if it is compounded continuously at 8%.

r PPlan In , replace P with 25,000, r with

0.08, A with $100,000, and then simplify.

continued

17.33t

Answer The account balance will reach $100,000 in approximately 17.33 years.

Substitute.

Divide.

Approximate using a calculator.

Execute1 100,000

ln0.08 25,000

continued

117.33 ln

0.08 25,000

1.3864 ln25,000

1.3864 ln ln 25,000A 1.3864 ln 25,000 ln A

11.513 ln A11.513e A

100,007.5 A

Because 17.33 was not the exact time, $100,007.45 is reasonable.

Use a calculator to approximate log 0.267 to four decimal places.

a) -1.3205

b) -0.5735

c) 1.3060

d) 1.8493

Use a calculator to approximate log 0.267 to four decimal places.

a) -1.3205

b) -0.5735

c) 1.3060

d) 1.8493

Use a calculator to approximate ln 21 to four decimal places.

a) 0.7636

b) 1.3316

c) 3.0445

d) 5.3688

Use a calculator to approximate ln 21 to four decimal places.

a) 0.7636

b) 1.3316

c) 3.0445

d) 5.3688

Exponential and Logarithmic Equations with Applications12.612.6

1. Solve equations that have variables as exponents.2. Solve equations containing logarithms.3. Solve applications involving exponential and

logarithmic functions. 4. Use the change-of-base formula.

Properties for Solving Exponential and Logarithmic Equations

For any real numbers b, x, and y, where b > 0 and b 1,1. If bx = by, then x = y. 2. If x = y, then bx = by. 3. For x > 0 and y > 0, if logbx = logby, then x = y.

4. For x > 0 and y > 0, if x = y, then logbx = logby.

5. For x > 0, if logbx = y, then by = x.

ExampleSolve 8x = 15.Solution

log8 log15x

Use if x = y, then logbx = logby (property 4).

1.3023x

Divide both sides by log 8.

The exact solution is . Using a calculator, we find .

Check 8x = 15

81.3023 = 15

15.0001 = 15

The answer is correct.

Solving Equations Containing LogarithmsTo solve equations containing logarithms, use the

properties of logarithms to simplify each side of the equation and then use one of the following.

If the simplification results in an equation in the form logbx = logby, use the fact that x = y, and then solve for the variable.

If the simplification results in an equation in the form logbx = y, write the equation in exponential form, by = x, and then solve for the variable (as we did in Section 10.3).

Example

Solution32 4x

2log ( 4) 3.x

This equation is in the form logbx = y, so write it in exponential form, by = x.

Solve for x.

ExampleSolveSolution

3 3 3log log ( 3) log 10.x x

Logarithms are defined for positive numbers only. A check will show that x = 5 the only solution.

3 3log ( 3) log 10x x ( 3) 10x x

2 3 10x x 2 3 10 0x x

5 2 0x x

5 0 or 2 0x x 5 or 2x x

ExampleSolve.Solution

ln( 4) ln( 1) ln(8 )x x x

20 8 12x x 4ln ln(8 )

(8 )( 1) 4x x x

28 8 4x x x x

29 8 4x x x

0 ( 6)( 2)x x

6 2x or x

These 2 solutions check and are the solutions for this equation.

Example

If $1000 is deposited into an account at 7.6% interest compounded continuously, how much money will be in the account after 8 years?

Understand We are given P = $1000, r = 0.076, and t = 8 and we are asked to find A.

Plan Use .rtA Pe

Execute 0.076(8)1000A e0.6081000A e

$1836.75A

continued

There will be $1836.75 in the account. Answer

Check1

Use the formula

1 1836.75ln .

0.076 1000t

8t It checks.

ExampleA nuclear reactor contains 200 grams of radioactive plutonium 239P. Plutonium disintegrates according to the formula A = A0e-0.0000284t. How much will remain after 5000 years?

Understand We are given A0 = 200 g and t = 5000 years and are asked to find A. Plan Use

Execute

0.00002840 .tA A e

0.0000284 5000200A e

173.52A

0.142200A e Simplify.

Substitute.

continued

About 173.52 grams will remain after 5000 years.

Answer

Check Use A = A0e-0.0000284t to verify that it takes 5000 years for 200 grams of 239P to disintegrate to 173.52 grams.

5000.9 tIt checks.

173.52 = 200e-0.0000284t

0.0000284173.52

0.0000284173.52ln ln

173.52ln 0.0000284

173.52ln

2000.0000284

Change-of-Base FormulaIn general, if a > 0, a 1, b > 0, b 1, and x > 0, then

In terms of common and natural logarithms,

log .log

log lnlog .

log lna

Example

Use the change-of-base formula to calculate log512. Round the answer to four decimal places.

Solution

log12log 12

1.5440

Check1.54405 12.0009 12 The answer is correct.

Solve log x + log(x – 3) = 1.

a) x = -2

b) x = 5

c) x = -2 or x = 5

d) x is undefined.

Solve log x + log(x – 3) = 1.

a) x = -2

b) x = 5

c) x = -2 or x = 5

d) x is undefined.

If $500 is deposited into an account at 9% interest compounded continuously, how much will be in the account after 5 years?

a) $653.79

b) $784.16

c) $892.36

d) $45,008.57

If $500 is deposited into an account at 9% interest compounded continuously, how much will be in the account after 5 years?

a) $653.79

b) $784.16

c) $892.36

d) $45,008.57

Use the change-of-base formula to approximate log756.

a) 0.48

b) 0.85

c) 2.07

Use the change-of-base formula to approximate log756.

a) 0.48

b) 0.85

c) 2.07

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