Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 11 - Cumulative Review - Page 670: 26b

Answer

$x=-3$

Work Step by Step

Using the properties of logarithms, then, \begin{array}{l} \log_4 \dfrac{1}{64}=x \\ \log_4 \dfrac{1}{4^3}=x \\ \log_4 4^{-3}=x \\ -3(\log_4 4)=x \\ -3(1)=x \\ x=-3 .\end{array}
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