Slide 4-1 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Slide 4-1Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Exponential and Logarithmic Functions

Chapter 4


4.1 Inverse Functions

Determine whether a function is one-to-one, and if it is, find a formula for its inverse.

Simplify expressions of the type (f f 1)(x) and (f 1 f)(x).


Inverses

When we go from an output of a function back to its input or inputs, we get an inverse relation. When that relation is a function, we have an inverse function.

Interchanging the first and second coordinates of each ordered pair in a relation produces the inverse relation.

Example: Consider the relation g given by g = {(2, 4), (3, 4), (8, 5). Solution: The inverse of the relation is {(4, 2), (4, 3), (5, 8)}.


Inverse Relation

If a relation is defined by an equation, interchanging the variables produces an equation of the inverse relation.

Example: Find an equation for the inverse of the relation: y = x2 2x.

Solution: We interchange x and y and obtain an equation of the inverse: x = y2 2y.

Graphs of a relation and its inverse are always reflections of each other across the line y = x.


Inverses of Functions

If the inverse of a function f is also a function, it is named f 1 and read “f-inverse.” The negative 1 in f 1 is not an exponent. This does not mean the reciprocal of f. f 1(x) is not equal to .1

( )f x


One-to-One Functions

A function f is one-to-one if different inputs have different outputs.

That is,

if a b

then

f(a) f(b).

A function f is one-to-one if when the outputs are the same, the inputs are the same.

That is,

if f(a) = f(b)

then

a = b.


Properties of One-to-One Functions and Inverses

If a function is one-to-one, then its inverse is a function. The domain of a one-to-one function f is the range of

the inverse f 1. The range of a one-to-one function f is the domain of

the inverse f 1. A function that is increasing over its domain or is

decreasing over its domain is a one-to-one function.


Horizontal-Line Test

If it is possible for a horizontal line to intersect the graph of a function more than once, then the function is not one-to-one and its inverse is not a function.



Graph f(x) = 3x + 4. Example: From the graph at the left, determine whether the function is one-to-one and thus has an inverse that is a function.

Solution: No horizontal line intersects the graph more than once, so the function is one-to-one. It has an inverse that is a function.

f(x) = 3x + 4



Graph f(x) = x2 2. Example: From the graph at the left, determine whether the function is one-to-one and thus has an inverse that is a function.

Solution: There are many horizontal lines that intersect the graph more than once. The inputs 1 and 1 have the same output, 1. Thus the function is not one-to-one. The inverse is not a function.


Obtaining a Formula for an Inverse

If a function f is one-to-one, a formula for its inverse can generally be found as follows: Replace f(x) with y. Interchange x and y. Solve for y. Replace y with f 1(x).


Example

Determine whether the function f(x) = 3x 2 is one-to-one, and if it is, find a formula for f 1(x).

Solution: The graph is that of a line and passes the horizontal-line test. Thus it is one-to-one and its inverse is a function.

1. Replace f(x) with y: y = 3x 2 2. Interchange x and y: x = 3y 2 3. Solve for y: x + 2 = 3y

4. Replace y with f 1(x):

f 1(x) =

2

3

xy

2

3

x


Example

Graph f(x) = 3x 2 and

f 1(x) =

using the same set of axes. Then compare the two graphs.

Solution: The solutions of the inverse function can be found from those of the original function by interchanging the first and second coordinates of each ordered pair. The graph f 1 is a reflection of the graph f across the line y = x.

2

3

x


Solution

73

42

20

51

f(x) = 3x 2x

24

11

02

15

f 1(x)

=

x

+ 2

3

x


Inverse Functions and Composition

If a function f is one-to-one, then f 1 is the unique function such that each of the following holds:

for each x in the domain of f, and

for each x in the domain of f 1.

1 1

1 1

( )( ) ( ( ))

( )( ) ( ( ))

f f x f f x x

f f x f f x x


Example

Given that f(x) = 7x 2, use composition of functions to show that f 1(x) = (x + 2)/7.

Solution: 1 1

1

( )( ) ( ( ))

(7 2)

(7 2) 2

77

7

f f x f f x

f x

x

x

x

1 1( )( ) ( ( ))

2( )

72

7( ) 272 2

f f x f f x

xf

x

x

x


Restricting a Domain

When the inverse of a function is not a function, the domain of the function can be restricted to allow the inverse to be a function. In such cases, it is convenient to consider “part” of the function by restricting the domain of f(x). If the domain is restricted, then its inverse is a function.


4.2 Exponential Functions

and Graphs Graph exponential equations and functions. Solve applied problems involving exponential

functions and their graphs.


Exponential Function

The function f(x) = ax, where x is a real number, a > 0 and a 1, is called the exponential function, base a.

The base needs to be positive in order to avoid the complex numbers that would occur by taking even roots of negative numbers.

Examples:

1( ) 3 ( ) ( ) (4.23)

3

xx xf x f x f x


Graphing Exponential Functions

To graph an exponential function, follow the steps listed:

1. Compute some function values and list the results

in a table.

2. Plot the points and connect them with a smooth curve. Be sure to plot enough points to

determine how steeply the curve rises.


Example

Graph the exponential function y = f(x) = 3x.

(3, 1/27)1/273

(2, 1/9)1/92

(1, 1/3)1/31

(3, 27)273

9

3

1

y = f(x) = 3x

(2, 9)2

(1, 3)1

(0, 1)0

(x, y)x


Example

Graph the exponential function . 1

( )3

x

y f x

(3, 1/27)1/273

(2, 1/9)1/92

(1, 1/3)1/31

(3, 27)273

9

3

1

(2, 9)2

(1, 3)1

(0, 1)0

(x, y)x 1( )

3

x

y f x


Example

Graph y = 3x + 2.

The graph is the graph of y = 3x shifted to left 2 units.

2433

812

271

90

3

1

1/3

y

1

2

3

x


Example

Graph y = 4 3x

The graph is a reflection of the graph of y = 3x across the x-axis, followed by a reflection across the y-axis and then a shift up of 4 units.

3.963

3.882

3.671

30

1

5

23

y

1

2

3

x


The Number e

e 2.7182818284…

Find each value of ex, to four decimal places, using the ex key on a calculator.

a) e4 b) e0.25

c) e2 d) e1

a) 54.5982 b) 0.7788 c) 7.3891 d) 0.3679


Graphs of Exponential Functions, Base e

Graph f(x) = ex.

7.3892

2.7181

1

0.368

0.135

f(x)

0

1

2

x


Example

Graph f(x) = 2 e3x.

1.992

1.951

1

18.09

401.43

f(x)

0

1

2

x


Example

Graph f(x) = ex+2.

0.1354

0.3683

20.0861

7.389

2.718

1

f(x)

0

1

2

x


4.3 Logarithmic Functions

and Graphs Graph logarithmic functions. Convert between exponential and logarithmic

equations. Find common and natural logarithms with and

without using a calculator. Change logarithm bases


Logarithmic Functions

These functions are inverses of exponential functions.

Graph: x = 3y.

1. Choose values for y.

2. Compute values for x.

3. Plot the points and connect them with a smooth curve.

* Note that the curve does not touch or cross the y-axis.


Logarithmic Functions continued

(1/27, 3)31/27

(1/9, 2)21/9

(1/3, 1)11/3

2

1

0

y

(9, 2)9

(3, 1)3

(1, 0)1

(x, y)x = 3y

Graph: x = 3y


Logarithmic Function, Base a

We define y = loga x as that number y such that x = ay, where x > 0 and a is a positive constant other than 1.

We read loga x as “the logarithm, base a, of x.”


Finding Certain Logarithms

Find each of the following logarithms. a) log2 16 b) log10 1000 c) log16 4 d) log10 0.001

a) The exponent to which we raise 2 to obtain 16 is 4; thus log2 16 = 4.b) The exponent to which we raise 10 to obtain 1000 is 3;

thus log10 1000 = 3.c) The exponent we raise 16 to get 4 is ½, so log16 4 = ½.d) We have The exponent to which we raise 10

to get 0.001 is 3, so log10 0.001 = 3.

33

1 110 .

1000 10


Logarithms

loga 1 = 0 and loga a = 1, for any logarithmic base a.

Convert each of the following to a logarithmic equation.

a) 25 = 5x

log5 25 = x

b) ew = 30

loge 30 = w

log A logarithm is an exponent!ayyx x a

The base remains the same.

The exponent is the logarithm.


Example

Convert each of the following to an exponential equation.

a) log7 343 = 3 log7 343 = 3 73 = 343

b) logb R = 12

logb R = 12 b12 = R

The logarithm is the exponent.

The base remains the same.


Example

Find each of the following common logarithms on a calculator. Round to four decimal places.

a) log 723,456

b) log 0.0000245

c) log (4)

Does not existERR: nonreal anslog (4)

4.61084.610833916log 0.0000245

5.85945.859412123log 723,456

RoundedReadoutFunction Value


Natural Logarithms

Logarithms, base e, are called natural logarithms. The abbreviation “ln” is generally used for natural logarithms. Thus,

ln x means loge x.

ln 1 = 0 and ln e = 1, for the logarithmic base e.


Example

Find each of the following natural logarithms on a calculator. Round to four decimal places.

a) ln 723,456

b) ln 0.0000245

c) ln (4)

Does not existERR: nonreal ansln (4)

10.616810.61683744ln 0.0000245

13.491813.49179501ln 723,456

RoundedReadoutFunction Value


Changing Logarithmic Bases

The Change-of-Base Formula

For any logarithmic bases a and b, and any positive number M,

loglog .

loga

ba

MM

b


Example

Find log6 8 using common logarithms.

Solution: First, we let a = 10, b = 6, and M = 8. Then we substitute into the change-of-base formula:

10

16

0

loglog

l

8

og

1.1606

68


Example

We can also use base e for a conversion.

Find log6 8 using natural logarithms.

Solution: Substituting e for a, 6 for b and 8 for M, we have

6 6

loglog

log

ln81.1606

ln 6

88 e

e


Graphs of Logarithmic Functions

Graph: y = f(x) = log6 x.

Select y. Compute x.

21/36

11/6

3216

236

16

01

yx,or 6y


Example

Graph each of the following. Describe how each graph can be obtained from the graph of y = ln x. Give the domain and the vertical asymptote of each function.

a) f(x) = ln (x 2) b) f(x) = 2 ln x c) f(x) = |ln (x + 1)|

14


Graph f(x) = ln (x 2)

The graph is a shift 2 units right. The domain is the set of all real numbers greater than 2. The line x = 2 is the vertical asymptote.

1.0995

0.6934

03

0.6932.5

1.3862.25

f(x)x


Graph f(x) = 2 ln x

The graph is a vertical shrinking, followed by a reflection across the x-axis, and then a translation up 2 units. The domain is the set of all positive real numbers. The y-axis is the vertical asymptote.

14

1.5985

1.7253

21

2.1730.5

2.5760.1

f(x)x


Graph f(x) = |ln (x + 1)|

The graph is a translation 1 unit to the left. Then the absolute value has the effect of reflecting negative outputs across the x-axis. The domain is the set of all real numbers greater than 1. The line x = 1 is the vertical asymptote.

1.9466

1.3863

0.6931

00

0.6930.5

f(x)x


Application: Walking Speed

In a study by psychologists Bornstein and Bornstein, it was found that the average walking speed w, in feet per second, of a person living in a city of population P, in thousands, is given by the function

w(P) = 0.37 ln P + 0.05.


Application: Walking Speed continued

The population of Philadelphia, Pennsylvania, is 1,517,600. Find the average walking speed of people living in Philadelphia.

Since 1,517,600 = 1517.6 thousand, we substitute 1517.6 for P, since P is in thousands: w(1517.6) = 0.37 ln 1517.6 + 0.05

2.8 ft/sec.The average walking speed of people living in Philadelphia is about 2.8 ft/sec.


4.4 Properties of Logarithmic

Functions Convert from logarithms of products, powers,

and quotients to expressions in terms of individual logarithms, and conversely. Simplify expressions of the type loga ax and .

loga xa


Logarithms of Products

The Product Rule

For any positive numbers M and N and any logarithmic base a,

loga MN = loga M + loga N.

(The logarithm of a product is the sum of the logarithms of the factors.)


Example

Express as a single logarithm: .

Solution:

23 3log logx w

2 23 3 3log log log ( )x w x w


Logarithms of Powers

The Power Rule

For any positive number M, any logarithmic base a, and any real number p,

loga Mp = p loga M.

(The logarithm of a power of M is the exponent times the logarithm of M.)


Examples

Express as a product. Express as a product.

3log 7a

3log 7 3log 7a a

5log 11a

1/ 55log 11 log 11

1log 11

5

a a

a


Logarithms of Quotients

The Quotient Rule

For any positive numbers M and N, and any logarithmic base a,

.

(The logarithm of a quotient is the logarithm of the numerator minus the logarithm of the denominator.)

log log loga a a

MM N

N


Examples

Express as a difference of logarithms.

Express as a single logarithm.

10log

log 10 log

a

a a

bb 125

log log 525w w

log 125 log 25w w


Applying the Properties

Express in terms of sums and differences of logarithms.

3 43 4 2

2

3 4 2

log log ( ) log

log log log

3log 4log 2log

a a a

a a a

a a a

w yw y z

z

w y z

w y z

3 4

2loga

w y

z


Example

Express as a single logarithm.

16log 2log log

3b b bx y z

6 2 1/ 3

61/ 3

2

6 1/ 3 6 3

2 2

16log 2log log log log log

3

log log

log , or log

b b b b b b

b b

b b

x y z x y z

xz

y

x z x z

y y


Final Properties

The Logarithm of a Base to a PowerFor any base a and any real number x,

loga ax = x.(The logarithm, base a, of a to a power is the power.)

A Base to a Logarithmic PowerFor any base a and any positive real number x,

(The number a raised to the power loga x is x.)

log .a xa x


Examples

Simplify.

a) loga a6

b) ln e8

Solution:

a) loga a6 = 6

b) ln e8 = 8

Simplify.

a)

b)

Solution:

a)

b)

7log7 w

7log7 w w

ln 8e

log 8ln 8 8ee e


4.5 Solving Exponential and Logarithmic Equations

Solve exponential and logarithmic equations.


Solving Exponential Equations

Equations with variables in the exponents, such as

3x = 40 and 53x = 25, are called exponential equations.

Base-Exponent Property

For any a > 0, a 1,

ax = ay x = y.


Example

Solve: .

Write each side with the same base.

Since the bases are the same number, 5, we can use the base-exponent property and set the exponents equal:

Check: 52x 3 = 125

52(3) 3 ? 125

53 ? 125

125 = 125 True

The solution is 3.

2 35 125x

2 3 35 5x

2 3 3

2 6

3

x

x

x


Graphical Solution

We will use the Intersect method. We graph y1 = and y2 = 53 2 35 x


Another Property

Property of Logarithmic Equality

For any M > 0, N > 0, a > 0, and a 1,

loga M = loga N M = N.

Solve: 2x = 50

log 2 log50

log 2 log50

log50

log 2

x

x

x

This is an exact answer. We cannot simplify further, but we can approximate using a calculator.

x 5.6439

We can check by finding 25.6439 50.


Graphical Solution

We will use the Intersect method. We graph y1 = 2x and y2 = 50


Example

Solve: e0.25w = 12

The solution is about 9.94.

0.25

0.25

12

ln ln12

0.25 ln12

ln12

0.259.93963

w

w

e

e

w

w

w


Solving Logarithmic Equations

Equations containing variables in logarithmic expressions, such as log2 x = 16 and log x + log (x + 4) = 1, are called logarithmic equations.

To solve logarithmic equations algebraically, we first try to obtain a single logarithmic expression on one side and then write an equivalent exponential equation.


Example

Solve: log4 x = 3 Check:

log4 x = 3

The solution is

4

3

3

log 3

4

1

41

64

x

x

x

x

14 64

34

log ? 3

log 4 ? 3

3 3

1.

64


Graphical Solution

We use the change of base formula and graph the equations

y1 =

y2 = 34

lnlog 3

ln 4

xx


Example

Solve: Check: For x = 3:

For x = 3:

2 2

2

3

2

2

log ( 1) log ( 1) 3

log ( 1)( 1) 3

( 1)( 1) 2

1 8

9

9

3

x x

x x

x x

x

x

x

x

2 2log ( 1) log ( 1) 3x x 2 2

2 2

2

2

log ( 1) log ( 1) 3

log 4 log 2 ? 3

log (4 2) ? 3

log 8 ? 3

3

3

3

3

2 2

2 2

log ( 1) log ( 1)? 3

log ( 2) log ( 4)? 3

3 3

Negative numbers do not have real-number logarithms. The solution is 3.


Example

Solve: ln ln( 4) ln3x x ln ln( 4) ln3

ln ln34

34

4 3( 4)4

3 12

12 2

6

x x

x

xx

xx

x xx

x x

x

x

The value 6 checks and is the solution.


4.6 Applications and Models:

Growth and Decay Solve applied problems involving exponential

growth and decay and compound interest. Find models involving exponential and

logarithmic functions.


Population Growth

The function P(t) = P0ekt, k > 0 can model many kinds of population growths.

In this function:

P0 = population at time 0,

P = population after time,

t = amount of time,

k = exponential growth rate.

The growth rate unit must be the same as the time unit.


Example

Population Growth of the United States. In 1990 the population in the United States was about 249 million and the exponential growth rate was 8% per decade. (Source: U.S. Census Bureau) Find the exponential growth function. What will the population be in 2020? After how long will the population be double what it

was in 1990?


Solution

At t = 0 (1990), the population was about 249 million. We substitute 249 for P0 and 0.08 for k to obtain the exponential growth function.

P(t) = 249e0.08t

In 2020, 3 decades later, t = 3. To find the population in 2020 we substitute 3 for t:

P(3) = 249e0.08(3) = 249e0.24 317.

The population will be approximately 317 million in 2020.


Solution continued

We are looking for the doubling time T.

498 = 249e0.08T

2 = e0.08T

ln 2 = ln e0.08T (Taking the natural logarithm on both sides)

ln 2 = 0.08T (ln ex = x)

= T

8.7 T

The population of the U.S. will double in about 8.7 decades or 87 years. This will be approximately in 2077.

ln 2

0.08


Interest Compound Continuously

The function P(t) = P0ekt can be used to calculate interest that is compounded continuously.

In this function:

P0 = amount of money invested,

P = balance of the account,

t = years,

k = interest rate compounded continuously.


Example

Suppose that $2000 is deposited into an IRA at an interest rate k, and grows to $5889.36 after 12 years.

What is the interest rate? Find the exponential growth function. What will the balance be after the first 5 years? How long did it take the $2000 to double?


Solution

At t = 0, P(0) = P0 = $2000. Thus the exponential growth function is P(t) = 2000ekt. We know that P(12) = $5889.36. We then substitute

and solve for k: $5889.36 = 2000e12k

The interest rate is about 9%.

12

12

5889.36

20005889.36

ln ln2000

5889.36ln 12

20005889.36

ln200012

0.09

k

k

e

e

k

k

k


Solution continued

The exponential growth function is

P(t) = 2000e0.09t.

The balance after 5 years is

P(5) = 2000e0.09(5)

= 2000e0.45

$3136.62


Solution continued

To find the doubling time T, set P(T) = 2 P0 = $4000 and solve for T.

4000 = 2000e0.09T

2 = e0.09T

ln 2 = ln e0.09T

ln 2 = 0.09T

= T

7.7 T

The original investment of $2000 doubled in about 7.7 years.

ln 2

0.09


Growth Rate and Doubling Time

The growth rate k and the doubling time T are related by

kT = ln 2

or

or

* The relationship between k and T does not depend on P0.

ln 2k

T

ln 2T

k


Example

A certain town’s population is doubling every 37.4 years. What is the exponential growth rate?

Solution: ln 2 ln 21.9%

37.4k

T


Models of Limited Growth

In previous examples, we have modeled population growth. However, in some populations, there can be factors that prevent a population from exceeding some limiting value.

One model of such growth is

which is called a logistic function. This function increases toward a limiting value a as t approaches infinity. Thus, y = a is the horizontal asymptote of the graph.

( )1 kt

aP t

be


Exponential Decay

Decay, or decline, of a population is represented by the function P(t) = P0ekt, k > 0.

In this function:

P0 = initial amount of the substance, P = amount of the substance left after time, t = time, k = decay rate.

The half-life is the amount of time it takes for half of an amount of substance to decay.


Graphs


Example

Carbon Dating. The radioactive element carbon-14 has a half-life of 5750 years. If a piece of charcoal that had lost 7.3% of its original amount of carbon, was discovered from an ancient campsite, how could the age of the charcoal be determined?

Solution: We know (from Example 5 in our book), that the function for carbon dating is

P(t) = P0e-0.00012t.

If the charcoal has lost 7.3% of its carbon-14 from its initial amount P0, then 92.7%P0 is the amount present.


Example continued

To find the age of the charcoal, we solve the equation for t :

The charcoal was about 632 years old.

0.000120 0

0.00012

0.00012

92.7%

0.927

ln 0.927 ln

ln 0.927 0.00012

ln 0.927

0.00012632

t

t

t

P P e

e

e

t

t

t

Documents

Slide 4-1 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley