College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Section 4.5 - Exponential and Logarithmic Equations - 4.5 Exercises - Page 447: 69


$x=3 $

Work Step by Step

$\bf{\text{Solution Outline:}}$ First, use the properties of logarithms to simplify the given logarithmic expression, $ \log_5(x+2)+\log_5(x-2)=1 ,$ to a single logarithm. Then convert to exponential form. Next, use concepts of quadratic equations to solve for the variable. Finally, it is a must to do checking at the end. $\bf{\text{Solution Details:}}$ Using $\log_b (xy)=\log_b x+\log_b y$ or the Product Rule of logarithms, the given equation is equivalent to \begin{array}{l}\require{cancel} \log_5[(x+2)(x-2)]=1.\end{array} Since $\log_by=x$ is equivalent to $y=b^x$, the equation above, in exponential form, is equivalent to \begin{array}{l}\require{cancel} (x+2)(x-2)=5^1 \\\\ (x+2)(x-2)=5 .\end{array} Using the properties of equality and the concepts of quadratic equations, the solution/s to the equation above is/are \begin{array}{l}\require{cancel} (x)^2-(2)^2=5 \\\\ x^2-4=5 \\\\ x^2=5+4 \\\\ x^2=9 \\\\ x=\pm\sqrt{9} \\\\ x=\pm3 .\end{array} Upon checking, only $ x=3 $ satisfies the original equation.
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