Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 9 - Section 9.5 - Exponential and Logarithmic Equations - Exercise Set - Page 727: 68

Answer

The solution set is $\{3\}$.

Work Step by Step

The given equation is $\log _2(x-1) +\log _2(x+1)=3$ Use product rule. $\log _2 [(x-1)(x+1)]=3$ Rewrite in exponential form. $(x-1)(x+1)= 2^3$ Use distributive property on the left. $x^2-x+x-1=8$ Subtract $8$ from both sides. $x^2-1-8=8-8$ Simplify. $x^2-9=0$ Factor. $(x-3)(x+3)=0$ Set both factors equal to $0$. $x-3=0$ or $x+3=0$ Isolate $x$. $x=3$ or $x=-3$ $-3$ is not in the domain of a logarithmic function. The solution is $x=3$.
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