Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 3 - Mid-Chapter Check Point - Page 477: 12

Answer

$$\frac{1}{2}$$

Work Step by Step

We simplify as follows: $$\log_{100}10= \log_{100}\sqrt{100}= \log_{100}100^{\frac{1}{2}}= \frac{1}{2} \log_{100}100=\frac{1}{2}(1)=\frac{1}{2}$$
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