Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 4 - Exponential and Logarithmic Functions - Section 4.5 Properties of Logarithms - 4.5 Assess Your Understanding - Page 331: 35



Work Step by Step

$\ln \sqrt[5]{6} = \ln 6^{\frac{1}{5}}$ $\because \log_a M^r = r \log_a M$ $\therefore \ln 6^{\frac{1}{5}} = \frac{1}{5} \ln{6}$ $\because \log_a (MN) = \log_a M+\log_a N$ $\therefore \ln6 = \ln(3 \cdot 2)= \ln3+\ln2$ With $\ln3 = b \hspace{20pt} \text{and} \hspace{20pt} \ln2=a$, then $\ln3+\ln2 = b+a$ $\ln6 = b+a$ Therfeore, $\ln \sqrt[5]{6} =\frac{[1}{5}\ln{6} = \boxed{\dfrac{1}{5}(b+a)}$
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