College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 6 - Review Exercises - Page 499: 18

Answer

$\log_2{\left(\frac{1}{8}\right)}=-3$

Work Step by Step

The given expression is equivalent to: \begin{align*} &=\log_2{\left(\frac{1}{2^3}\right)}\\\\ &=\log_2{\left(2^{-3}\right)}\\\\ \end{align*} Using the rule $\log_a{a^b}=b$, then $\log_2{\left(2^{-3}\right)}=-3$ Thus, $\log_2{\left(\frac{1}{8}\right)}=-3$.
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