Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 1 - Section 1.5 - Inverse Functions and Logarithms - 1.5 Exercises - Page 67: 37


a) $\log_{10}{40}+\log_{10}{2.5}=2$ b) $\log_8{60}-\log_8{3}-\log_85=\frac{2}{3}$

Work Step by Step

a) $\log_{10}{40}+\log_{10}{2.5}=\log_{10}=\log_{10}{(40\times 2.5)}=\log_{10}{100}=\log_{10}{10^2}=2\log_{10}{10}=2$ b) $\log_8{60}-\log_8{3}-\log_85=\log_8{(60\div3\div5)}=\log_84=\log_82^2=2\log_82=2\log_88^\frac{1}{3}=\frac{2}{3}\log_88=\frac{2}{3}$
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