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: 93

Answer

$\frac{3}{2}u-\frac{5}{2}v$

Work Step by Step

$ln(\sqrt {\frac{a^3}{b^5}})=\frac{3}{2}ln\ a-\frac{5}{2}ln\ b=\frac{3}{2}u-\frac{5}{2}v$
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