Answer
$x_{1}=1$ 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 $x_{2}=-8$
Algebra: A Combined Approach (4th Edition)
by Martin-Gay, Elayn
Published by
Pearson
ISBN 10:
0321726391
ISBN 13:
978-0-32172-639-1
Chapter 12 - Review - Page 906: 52
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