Answer
True.
Work Step by Step
Please note that we have the following relation:$$\log_bM^p=p\log_bM.$$So we have$$\ln \sqrt{2}=\ln 2^{\frac{1}{2}}=\frac{1}{2}\ln 2= \frac{\ln 2}{2}$$
Precalculus (6th Edition) Blitzer
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-13446-914-3
ISBN 13:
978-0-13446-914-0
Chapter 3 - Section 3.3 - Properties of Logarithms - Exercise Set - Page 477: 125
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