Loading in 5 sec....

Chapter 5: Exponential and Logarithmic FunctionsPowerPoint Presentation

Chapter 5: Exponential and Logarithmic Functions

- By
**phuc** - Follow User

- 134 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' Chapter 5: Exponential and Logarithmic Functions' - phuc

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

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 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 Compounding Problem

- 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.

5.3 Additional Properties of Logarithms Compounding Problem

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 Compounding Problem

Example Assume all variables are positive. Use the

properties of logarithms to rewrite the expression

Solution

5.3 Example Using Logarithm Properties Compounding Problem

Example Use the properties of logarithms to write

as a single logarithm

with coefficient 1.

Solution

5.3 The Change-of-Base Rule Compounding Problem

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 Compounding Problem

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 Compounding Problem

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 Compounding Problem

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.

Download Presentation

Connecting to Server..