Chapter 5: Exponential and Logarithmic Functions

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# Chapter 5: Exponential and Logarithmic Functions - PowerPoint PPT Presentation

Chapter 5: Exponential and Logarithmic Functions. 5.1 Inverse Functions 5.2 Exponential Functions 5.3 Logarithms and Their Properties 5.4 Logarithmic Functions 5.5 Exponential and Logarithmic Equations and Inequalities

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Chapter 5: Exponential and Logarithmic Functions

5.1 Inverse Functions

5.2 Exponential Functions

5.3 Logarithms and Their Properties

5.4 Logarithmic Functions

5.5 Exponential and Logarithmic Equations and Inequalities

5.6 Further Applications and Modeling with Exponential and Logarithmic Functions

5.3 Logarithms and Their Properties

Logarithm

For all positive numbers a, where a 1,

A logarithm is an exponent, and loga x is the exponent to which a must be raised in order to obtain x. The number a is called the base of the logarithm, and x is called the argument of the expression loga x. The value of x will always be positive.

5.3 Examples of Logarithms

Exponential Form Logarithmic Form

Example Solve

Solution

5.3 Solving Logarithmic Equations

Example Solve a)

Solution

Since the base must

be positive, x = 2.

5.3 The Common Logarithm – Base 10

For all positive numbers x,

Example Evaluate

Solution Use a calculator.

5.3 Application of the Common Logarithm

Example In chemistry, the pH of a solution is defined as

where [H3O+] is the hydronium ion

concentration in moles per liter. The pH value is a measure of

acidity or alkalinity of a solution. Pure water has a pH of 7.0,

substances with a pH greater than 7.0 are alkaline, and those

less than 7.0 are acidic.

• Find the pH of a solution with [H3O+] = 2.5×10-4.
• Find the hydronium ion concentration of a solution with pH = 7.1.

Solution

• pH = –log [H3O+] = –log [2.5×10-4]  3.6
• 7.1 = –log [H3O+]  –7.1 = log [H3O+]

 [H3O+] = 10-7.1  7.9 ×10-8

5.3 The Natural Logarithm – Base e

For all positive numbers x,

• On the calculator, the natural logarithm key is usually found in conjunction with the e x key.
5.3 Using Natural Logarithms to Solve a Continuous Compounding Problem

Example Suppose that \$1000 is invested at 3%

annual interest, compounded continuously. How

long will it take for the amount to grow to \$1500?

Analytic Solution

5.3 Using Natural Logarithms to Solve a Continuous Compounding Problem

Graphing Calculator Solution

Let Y1 = 1000e.03t and Y2 = 1500.

The table shows that when time (X) is 13.5 years, the

amount (Y1) is 1499.3  1500.

5.3 Properties of Logarithms
• For a > 0, a  1, and any real number k,
• loga 1 = 0,
• loga ak = k,
• a logak = k, k > 0.

Property 1 is true because a0 = 1 for any value of a.

Property 2 is true since in exponential form:

Property 3 is true since logak is the exponent to which a must be raised in order to obtain k.

For x > 0, y > 0, a > 0, a 1, and any real number r,

Product Rule

Quotient Rule

Power Rule

Examples Assume all variables are positive. Rewrite each

expression using the properties of logarithms.

5.3 Example Using Logarithm Properties

Example Assume all variables are positive. Use the

properties of logarithms to rewrite the expression

Solution

5.3 Example Using Logarithm Properties

Example Use the properties of logarithms to write

as a single logarithm

with coefficient 1.

Solution

5.3 The Change-of-Base Rule

Change-of-Base Rule

For any positive real numbers x, a, and b, where

a 1 and b 1,

Proof Let

5.3 Using the Change-of-Base Rule

Example Evaluate each expression and round to

four decimal places.

Solution Note in the figures below that using

either natural or common logarithms produce the

same results.

5.3 Modeling the Diversity of Species

Example One measure of the diversity of species

in an ecological community is the index of diversity,

where

and P1, P2, . . . , Pn are the proportions of a sample

belonging to each of n species found in the sample.

Find the index of diversity in a community where

there are two species, with 90 of one species and 10

of the other.

5.3 Modeling the Diversity of Species

Solution Since there are a total of 100 members in

the community, P1 = 90/100 = .9, and P2 = 10/100 = .1.

Interpretation of this index varies. If two species are

equally distributed, the measure of diversity is 1. If

there is little diversity, H is close to 0. In this case

H .5, so there is neither great nor little diversity.
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