College Algebra (11th Edition)

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

Chapter 4 - Test: 9

Answer

$\log_{9} 13\approx1.1674$

Work Step by Step

By using the change of base formula, we can rewrite the given logarithm as: $\log_{9} 13=\frac{\log_{10} 13}{\log_{10} 9}$ This can be easily calculated: $\frac{\log_{10} 13}{\log_{10} 9}\approx1.1674$
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