Intermediate Algebra (6th Edition)

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

Chapter 9 - Test - Page 599: 16

Answer

$\approx1.0686$

Work Step by Step

Using the Change-of-Base Formula, $\log_b a=\dfrac{\log a}{\log b}$, then, \begin{array}{l} \log_7 8 \\\\= \dfrac{\log 8}{\log 7} \\\\ \approx1.0686 .\end{array}
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