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: 94

Answer

$\frac{1}{3}u+4v$

Work Step by Step

$ln(\sqrt[3] a\cdot b^4)=\frac{1}{3}ln\ a+4ln\ b=\frac{1}{3}u+4v$
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