Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.4 Evaluation Logarithms and the Change-of-Base Theorem - 4.4 Exercises - Page 460: 81

Answer

$-0.2537$

Work Step by Step

Use a calculator and the change-of-base theorem, $log_{8}0.59=\frac{log0.59}{log8}=\frac{ln0.59}{ln8}\approx-0.2537$
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