Slide 4.3- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

OBJECTIVES

Logarithmic Function

Learn to define logarithmic functions.

Learn to evaluate logarithms.

Learn to find the domains of logarithmic functions.

Learn to graph logarithmic functions.

Learn to use logarithms to evaluate exponential equations.

SECTION 4.3

1

2

3

4

5

Slide 4.3- 3 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

DEFINITION OF THELOGARITHMIC FUNCTION

For x > 0, a > 0, and a ≠ 1,

y loga x if and only if x ay .

The function f (x) = loga x, is called the logarithmic function with base a.

The logarithmic function is the inverse function of the exponential function.

Slide 4.3- 4 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 1Converting from Exponential to Logarithmic Form

a. 43 64

Write each exponential equation in logarithmic form.

b. 1

2

4

1

16c. a 2 7

Solution

a. 43 64 log4 64 3

b. 1

2

4

1

16 log1 2

1

164

c. a 2 7 loga 7 2

Slide 4.3- 5 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 2Converting from Logarithmic Form to Exponential Form

a. log3 243 5

Write each logarithmic equation in exponential form.

b. log2 5 x c. loga N x

Solution

a. log3 243 5 243 35

b. log2 5 x 5 2x

c. loga N x N ax

Slide 4.3- 6 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 3 Evaluating Logarithms

a. log5 25

Find the value of each of the following logarithms.

b. log2 16 c. log1 3 9

d. log7 7 e. log6 1 f. log4

1

2

Solution

a. log5 25 y 25 5y or 52 5y y 2

b. log2 16 y 16 2y or 24 2y y 4

Slide 4.3- 7 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 3 Evaluating Logarithms

Solution continued

d. log7 7 y 7 7y or 71 7y y 1

e. log6 1 y 1 6y or 60 6y y 0

f. log4

1

2y

1

24 y or 2 1 22 y y

1

2

c. log1 3 9 y 9 1

3

y

or 32 3 y y 2

Slide 4.3- 8 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 4 Using the Definition of Logarithm

a. log5 x 3

Solve each equation.

b. log3

1

27y

c. logz 1000 3 d. log2 x2 6x 10 1

Solutiona. log5 x 3

x 5 3

x 1

53 1

125The solution set is

1

125

.

Slide 4.3- 9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 4 Using the Definition of Logarithm

Solution continued

b. log3

1

27y

1

273y

3 3 3y

3 y

c. logz 1000 3

1000 z3

103 z3

10 z

The solution set is 3 .

The solution set is 10 .

Slide 4.3- 10 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 4 Using the Definition of Logarithm

Solution continuedd. log2 x2 6x 10 1

x2 6x 10 21 2

x2 6x 8 0

x 2 x 4 0

x 2 0 or x 4 0

x 2 or x 4

The solution set is 2, 4 .

Slide 4.3- 11 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

DOMAIN OF LOGARITHMIC FUNCTION

Domain of f –1(x) = loga x is (0, ∞)

Range of f –1(x) = loga x is (–∞, ∞)

Recall thatDomain of f (x) = ax is (–∞, ∞)Range of f (x) = ax is (0, ∞)

Since the logarithmic function is the inverse of the exponential function,

Thus, the logarithms of 0 and negative numbers are not defined.

Slide 4.3- 12 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 4 Finding the Domain

a. f x log3 2 x Find the domain of each function.

2 x 0

2 x

b. f x log3

x 2

x 1

Solution

a. Domain of a logarithmic function must be positive, that is,

The domain of f is (–∞, 2).

Slide 4.3- 13 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 4 Finding the Domain

Solution continuedx 2

x 1 0b. Domain must be positive, that is,

The domain of f is (–∞, –1) U (2, ∞).

Set numerator = 0 and denominator = 0.x – 2 = 0 x + 1 = 0x = 2 x = –1

Create a sign graph.

Slide 4.3- 14 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

BASIC PROPERTIES OF LOGARITHMS

1. loga a = 1.

2. loga 1 = 0.

3. loga ax = x, for any real number x.4. aloga x x, for any x 0.

For any base a > 0, with a ≠ 1,

Slide 4.3- 15 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 6 Sketching a Graph

Sketch the graph of y = log3 x.Solution by plotting points.

Make a table of values.

x y = log3 x (x, y)

3–3 = 1/27 –3 (1/27, –3)

3–2 = 1/9 –2 (1/9, –2)

3–3 = 1/3 –1 (1/3, –1)

30 = 1 0 (1, 0)

31 = 3 1 (3, 1)

32 = 9 2 (9, 2)

Slide 4.3- 16 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 6 Sketching a Graph

Solution continued

Plot the ordered pairs and connect with a smooth curve to obtain the graph of y = log3 x.

Slide 4.3- 17 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 6 Sketching a Graph

Solution by using the inverse function

Graph y = f (x) = 3x

Reflect the graph of y = 3x in the line y = x to obtain the graph of y = f –1(x) = log3 x

Slide 4.3- 18 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

PROPERTIES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS

Exponential Function f (x) = ax

Logarithmic Function f (x) = loga x

Domain (0, ∞) Range (–∞, ∞)

Domain (–∞, ∞) Range (0, ∞)

x-intercept is 1 No y-intercept

y-intercept is 1 No x-intercept

x-axis (y = 0) is the horizontal asymptote

y-axis (x = 0) is the vertical asymptote

Slide 4.3- 19 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

PROPERTIES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS

Exponential Function f (x) = ax

Logarithmic Function f (x) = loga x

Is one-to-one, that is, logau = logav if and only if u = v

Is one-to-one , that is, au = av if and only if u = v

Increasing if a > 1 Decreasing if 0 < a < 1

Increasing if a > 1 Decreasing if 0 < a < 1

Slide 4.3- 20 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

GRAPHS OF LOGARITHMIC FUNCTIONS

f (x) = loga x (0 < a < 1)f (x) = loga x (a > 1)

Slide 4.3- 21 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 7 Using Transformations

Start with the graph of f (x) = log3 x and use transformations to sketch the graph of each function.

a. f x log3 x 2

c. f x log3 x

b. f x log3 x 1

d. f x log3 x

State the domain and range and the vertical asymptote for the graph of each function.

Slide 4.3- 22 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 7 Using Transformations

Solution

Shift up 2Domain (0, ∞)Range (–∞, ∞)Vertical asymptote x = 0

a. f x log3 x 2

Slide 4.3- 23 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 7 Using Transformations

Solution continued

Shift right 1Domain (1, ∞)Range (–∞, ∞)Vertical asymptote x = 1

b. f x log3 x 1

Slide 4.3- 24 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 7 Using Transformations

Solution continued

Reflect graph of y = log3 x in the x-axis Domain (0, ∞)Range (–∞, ∞)Vertical asymptote x = 0

c. f x log3 x

Slide 4.3- 25 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 7 Using Transformations

Solution continued

Reflect graph of y = log3 x in the y-axis Domain (∞, 0)Range (–∞, ∞)Vertical asymptote x = 0

d. f x log3 x

Slide 4.3- 26 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

COMMON LOGARITHMS

1. log 10 = 1.

2. log 1 = 0.

3. log 10x = x4. 10log x x

The logarithm with base 10 is called the common logarithm and is denoted by omitting the base: log x = log10 x. Thus,

y = log x if and only if x = 10 y.

Applying the basic properties of logarithms

Slide 4.3- 27 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 8 Using Transformations to Sketch a Graph

Sketch the graph of y 2 log x 2 .Solution

Graph f (x) = log x and shift it right 2 units.

f x log x f x log x 2

Slide 4.3- 28 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 8 Using Transformations to Sketch a Graph

Solution continued

y log x 2 Reflect in x-axis

y 2 log x 2 Shift up 2

Slide 4.3- 29 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

NATURAL LOGARITHMS

1. ln e = 1

2. ln 1 = 0

3. log ex = x4. eln x x

The logarithm with base e is called the natural logarithm and is denoted by ln x. That is, ln x = loge x. Thus,

y = ln x if and only if x = e y.

Applying the basic properties of logarithms

Slide 4.3- 30 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 9 Evaluating the Natural Logarithm

Evaluate each expression.

a. ln e4 b. ln1

e2.5 c. ln 3

Solution

a. ln e4 4

b. ln1

e2.5 ln e 2.5 2.5

(Use a calculator.)c. ln 3 1.0986123

Slide 4.3- 31 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

NEWTON’S LAW OF COOLING

Newton’s Law of Cooling states that

where T is the temperature of the object at time t, Ts is the surrounding temperature, and T0 is the value of T at t = 0.

T Ts T0 Ts e kt ,

Slide 4.3- 32 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 10 McDonald’s Hot Coffee

The local McDonald’s franchise has discovered that when coffee is poured from a pot whose contents are at 180ºF into a noninsulated pot in the store where the air temperature is 72ºF, after 1 minute the coffee has cooled to 165ºF. How long should the employees wait before pouring the coffee from this noninsulated pot into cups to deliver it to customers at 125ºF?

Slide 4.3- 33 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 10 McDonald’s Hot Coffee

Use Newton’s Law of Cooling with T0 = 180 and Ts = 72 to obtain

Solution

We have T = 165 and t = 1.

T 72 180 72 e kt

T 72 108e kt

165 72 108e k

93

108e k

ln93

108

k

k 0.1495317

Slide 4.3- 34 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 10 McDonald’s Hot Coffee

Substitute this value for k.

Solution continued

Solve for t when T = 125.

T 72 108e 0.1495317t

125 72 108e 0.1495317t

125 72

108e 0.1495317t

ln53

108

0.1495317t

t 1

0.1495317ln

53

108

t 4.76

The employee should wait about 5 minutes.

Slide 4.3- 35 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

GROWTH AND DECAY MODEL

A is the quantity after time t.A0 is the initial (original) quantity (when t = 0).r is the growth or decay rate per period.t is the time elapsed from t = 0.

A A0ert

Slide 4.3- 36 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 11 Chemical Toxins in a Lake

In a large lake, one-fifth of the water is replaced by clean water each year. A chemical spill deposits 60,000 cubic meters of soluble toxic waste into the lake.

a. How much of this toxin will be left in the lake after four years?

b. How long will it take for the toxic chemical in the lake to be reduced to 6000 cubic meters?

Slide 4.3- 37 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 11 Chemical Toxins in a Lake

A A0ert

A 60,000e

1

5

t

One-fifth (1/5) of the water in the lake is replaced by clean water every year, the decay rate for the toxin is r = –1/5 and A0 = 60,000. So,

Solution

where A is the amount of toxin (in cubic meters) after t years.

Slide 4.3- 38 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 11 Chemical Toxins in a Lake

A 60,000e

1

5

4

A 26,959.74 cm3

a. Substitute t = 4.

Solution continued

b. Substitute A = 6000 and solve for t.

6000 60,000e

1

5

t

1

10e

1

5

t

0.1 e

1

5

t

ln 0.1 1

5t

t 5 ln 0.1 t 11.51 years