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Slide 1:

Copyright © Cengage Learning. All rights reserved. 3 Exponential and Logarithmic Functions

Slide 2:

3.4 EXPONENTIAL AND LOGARITHMIC EQUATIONS Copyright © Cengage Learning. All rights reserved.

What You Should Learn:

Solve simple exponential and logarithmic equations. Solve more complicated exponential equations. Solve more complicated logarithmic equations. Use exponential and logarithmic equations to model and solve real-life problems. What You Should Learn

Slide 4:

Introduction

Introduction:

Introduction There are two basic strategies for solving exponential or logarithmic equations. The first is based on the One-to-One Properties and was used to solve simple exponential and logarithmic equations. The second is based on the Inverse Properties. For a > 0 and a ≠ 1, the following properties are true for all x and y for which log a x and log a y are defined.

Introduction:

Introduction One-to-One Properties a x = a y if and only if x = y . log a x = log a y if and only if x = y . Inverse Properties a log a x = x log a a x = x

Example 1 – Solving Simple Equations:

Example 1 – Solving Simple Equations Original Rewritten Equation Equation Solution Property a. 2 x = 32 2 x = 2 5 x = 5 One-to-One b. ln x – ln 3 = 0 ln x = ln 3 x = 3 One-to-One c. = 9 3 – x = 3 2 x = –2 One-to-One d. e x = 7 ln e x = ln 7 x = ln 7 Inverse e. ln x = –3 e ln x = e – 3 x = e –3 Inverse f. log x = – 1 10 log x = 10 – 1 x = 10 – 1 = Inverse g. log 3 x = 4 3 log 3 x = 3 4 x = 81 Inverse

Introduction:

Introduction The strategies used in Example 1 are summarized as follows.

Slide 9:

Solving Exponential Equations

Example 2 – Solving Exponential Equations:

Example 2 – Solving Exponential Equations Solve each equation and approximate the result to three decimal places, if necessary. a. e – x 2 = e – 3 x – 4 b. 3(2 x ) = 42

Example 2(a) – Solution:

Example 2(a) – Solution e – x 2 = e – 3 x – 4 – x 2 = –3 x – 4 x 2 – 3 x – 4 = 0 ( x + 1)( x – 4) = 0 ( x + 1) = 0 ( x – 4) = 0 The solutions are x = –1 and x = 4. Check these in the original equation. Set 1st factor equal to 0. Set 2nd factor equal to 0. Write original equation. One-to-One Property Write in general form. Factor. x = 4 x = –1

Example 2(b) – Solution:

Example 2(b) – Solution 3(2 x ) = 42 2 x = 14 log 2 2 x = log 2 14 x = log 2 14 x =  3.807 The solution is x = log 2 14  3.807. Check this in the original equation. cont’d Write original equation. Divide each side by 3. Take log (base 2) of each side. Inverse Property Change-of-base formula

Slide 13:

Solving Logarithmic Equations

Solving Logarithmic Equations:

Solving Logarithmic Equations To solve a logarithmic equation, you can write it in exponential form. ln x = 3 e ln x = e 3 x = e 3 This procedure is called exponentiating each side of an equation. Logarithmic form Exponentiate each side. Exponential form

Example 6 – Solving Logarithmic Equations:

Example 6 – Solving Logarithmic Equations a. ln x = 2 e ln x = e 2 x = e 2 b. log 3 (5 x – 1) = log 3 ( x + 7) 5 x – 1 = x + 7 4 x = 8 x = 2 Original equation Exponentiate each side. Inverse Property Original equation One-to-One Property Add – x and 1 to each side. Divide each side by 4.

Example 6 – Solving Logarithmic Equations:

Example 6 – Solving Logarithmic Equations c. log 6 (3 x + 14) – log 6 5 = log 6 2 x 3 x + 14 = 10 x –7 x = –14 x = 2 Quotient Property of Logarithms Original equation One-to-One Property Cross multiply. Isolate x . Divide each side by –7. cont’d

Slide 17:

Applications

Example 10 – Doubling an Investment:

Example 10 – Doubling an Investment You have deposited $500 in an account that pays 6.75% interest, compounded continuously. How long will it take your money to double? Solution: Using the formula for continuous compounding, you can find that the balance in the account is A = Pe rt A = 500 e 0.0675 t .

Example 10 – Solution:

Example 10 – Solution To find the time required for the balance to double, let A = 1000 and solve the resulting equation for t . 500 e 0.0675 t = 1000 e 0.0675 t = 2 ln e 0.0675 t = ln 2 0.0675 t = ln 2 t = t  10.27 cont’d Let A = 1000. Take natural log of each side. Divide each side by 500. Inverse Property Divide each side by 0.0675. Use a calculator.

Example 10 – Solution:

Example 10 – Solution The balance in the account will double after approximately 10.27 years. This result is demonstrated graphically in Figure 3.31. cont’d Figure 3.31

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