Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 9 - Review Exercises - Page 638: 42

Answer

$1.7925$

Work Step by Step

Using the Change-of-Base Formula, the given expression, $ \log_4 12 ,$ is equivalent to \begin{align*}\require{cancel} & \dfrac{\log_{10} 12}{\log_{10} 4} \\\\&= \dfrac{\log 12}{\log 4} \\\\&\approx 1.7925 .\end{align*} Hence, the expression $ \log_4 12 $ is approximately equal to $ 1.7925 $.
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