Answer
-1
Work Step by Step
If $b\gt0$ and $b\ne1$, then $y=log_{b}x$ means $x=b^{y}$ for every $x\gt0$ and every real number $y$.
Therefore, $log_{8}(8)^{-1}=-1$, because $8^{-1}=(8)^{-1}$.
Intermediate Algebra (6th Edition)
by Martin-Gay, Elayn
Published by
Pearson
ISBN 10:
0321785045
ISBN 13:
978-0-32178-504-6
Chapter 9 - Section 9.5 - Logarithmic Functions - Exercise Set - Page 572: 75
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