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 - Page 896: 24



Work Step by Step

$\log_{3}x+\log_{3}(x+6)=3$ Combine $\log_{3}x+\log_{3}(x+6)$ as the $\log$ of a product: $\log_{3}x(x+6)=3$ $\log_{3}(x^{2}+6x)=3$ Rewrite in exponential form: $3^{3}=x^{2}+6x$ $x^{2}+6x=27$ $x^{2}+6x-27=0$ Solve this equation by factoring: $(x+9)(x-3)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x+9=0$ $x=-9$ $x-3=0$ $x=3$ The initial equation is undefined for $x=-9$, so the answer is just $x=3$
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