Exponential and Logarithmic Equations (Day 1) 3.4

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Presentation transcript:

Exponential and Logarithmic Equations (Day 1) 3.4 Copyright © Cengage Learning. All rights reserved.

Objectives Solve simple exponential and logarithmic equations. Solve more complicated exponential equations.

Introduction

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

Introduction One-to-One Properties ax = ay if and only if x = y. loga x = loga y if and only if x = y. Inverse Properties aloga x = x loga ax = x *Can also use Mr. C’s “Circular Rule of Three”

Example 1 – Solving Simple Equations Original Rewritten Equation Equation Solution Property a. 2x = 32 One-to-One b. ln x – ln 3 = 0 One-to-One c. = 9 One-to-One d. ex = 7 Inverse/CR3 e. ln x = –3 Inverse/CR3 f. log x = –1 Inverse/CR3 g. log3 x = 4 Inverse/CR3

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

Solving Exponential Equations

Example 2 – Solving Exponential Equations Solve each equation and approximate the result to three decimal places, if necessary. a. e – x2 = e – 3x – 4 b. 3(2x) = 42

3.4 Example – Worked Solutions

Example 1 – Solving Simple Equations Original Rewritten Equation Equation Solution Property a. 2x = 32 2x = 25 x = 5 One-to-One b. ln x – ln 3 = 0 ln x = ln 3 x = 3 One-to-One c. = 9 3– x = 32 x = –2 One-to-One d. ex = 7 ln ex = ln 7 x = ln 7 Inverse e. ln x = –3 eln x = e– 3 x = e–3 Inverse f. log x = –1 10log x = 10–1 x = 10–1 = Inverse g. log3 x = 4 3log3 x = 34 x = 81 Inverse

Example 2 – Solving Exponential Equations Solve each equation and approximate the result to three decimal places, if necessary. a. e – x2 = e – 3x – 4 b. 3(2x) = 42

Example 2(a) – Solution e –x2 = e – 3x – 4 –x2 = –3x – 4 x2 – 3x – 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. Write original equation. One-to-One Property Write in general form. Factor. x = –1 Set 1st factor equal to 0. x = 4 Set 2nd factor equal to 0.

Example 2(b) – Solution cont’d 3(2x) = 42 2x = 14 log2 2x = log2 14 x = log2 14 x = ≈3.807 The solution is x = log2 14 ≈ 3.807. Check this in the original equation. Write original equation. Divide each side by 3. Take log (base 2) of each side. Inverse Property Change-of-base formula