Algebra: A Combined Approach (4th Edition)

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

Chapter 12 - Review: 52


$x_{1}=1$ and $x_{2}=-8$

Work Step by Step

$\log_{8}(x^2+7x) = 1$ $8^{\log_{8}(x^2+7x)} = 8^1$ $x^2+7x = 8$ $x^2+7x-8=0$ $(x-1)(x+8)=0$ $x_{1}=1$ and $x_{2}=-8$
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