Precalculus (6th Edition) Blitzer

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

Chapter 3 - Section 3.2 - Exponential Functions - Exercise Set - Page 466: 143



Work Step by Step

As we know, $\log_a1=0$, $\log_aa=1$, and $\log_ab^c=c\log_ab$. So we have$$\log_381=\log_33^4=4\log_33=4(1)=4, \quad \log_{\pi }1=0, \\ \log_{2\sqrt{2}}8=\log_{\sqrt{8}}(\sqrt{8})^2=2\log_{\sqrt{8}} \sqrt{8}=2(1)=2, \quad \log 0.001= \log_{10}10^{-3}=-3\log_{10}10=-3(1)=-3$$Thus,$$\frac{\log_381-\log_{\pi }1}{\log_{2\sqrt{2}}8-\log 0.001}= \frac{4-0}{2-(-3)}= \frac{4}{5}=0.8$$
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