College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Section 4.4 - Evaluating Logarithms and the Change-of-Base Theorem - 4.4 Exercises - Page 438: 87

Answer

$\approx1.9376$

Work Step by Step

Using $\log_a b=\dfrac{\log b}{\log a}$ or the Change-Of-Base Theorem, the given expression, $ \log_{\sqrt{13}} 12 ,$ evaluates to \begin{array}{l}\require{cancel} \dfrac{\log 12}{\log \sqrt{13}} \\\\ \approx1.9376 .\end{array}
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