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: 19



Work Step by Step

We simplify as follows: $$\log_{100}0.1=\log_{100}10^{-1}=\log_{100}(100^{\frac{1}{2}})^{-1}= \log_{100}100^{-\frac{1}{2}}=-\frac{1}{2}\log_{100}100=-\frac{1}{2}(1)=-\frac{1}{2}$$
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