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: 32

Answer

$-a$

Work Step by Step

With $0.5=\frac{1}{2}$, $\ln0.5 = \ln \left(\dfrac{1}{2} \right)$ Recall: $\ln\left(\dfrac{M}{N}\right) = \ln M-\ln N$ Usin g the rule above gives: $\ln \left(\dfrac{1}{2} \right) = \ln1-\ln2$ With $\ln1 =0 \hspace{20pt} \text{and} \hspace{20pt} \ln2=a$, then $\ln1-\ln2 = 0-a=-a$ Therefore, $\ln0.5 = \boxed{-a}$
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