Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 12 - Section 12.8 - Exponential and Logarithmic Equations and Problem Solving - Exercise Set: 30

Answer

$x=9$

Work Step by Step

$\log_{3}x+\log_{3}(x-8)=2$ Combine $\log_{3}x+\log_{3}(x-8)$ as a the $\log$ of a product: $\log_{3}x(x-8)=2$ $\log_{3}(x^{2}-8x)=2$ Rewrite in exponential form: $3^{2}=x^{2}-8x$ $x^{2}-8x=9$ $x^{2}-8x-9=0$ Solve this equation by factoring: $(x+1)(x-9)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x+1=0$ $x=-1$ $x-9=0$ $x=9$ The original equation is undefined for $x=-1$, the answer is just $x=9$
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