Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.4* General Logarithmic and Exponential Functions - 6.4* Exercise - Page 463: 9


(a) $log_{10}40-log_{10}2.5=2$ (b) $log_{8}60-log_{8}3-log_{8}5=\frac{2}{3}$

Work Step by Step

(a) Use logarithmic properties, $logx+logy=log(xy)$, $y logx=logx^{y}$ and $log x-log y=log(\frac{x}{y})$ $log_{10}40-log_{10}2.5=log_{10}(40)(2.5)$ $=log_{10}100$ $=\log_{10}(10^{2})$ Hence, $log_{10}40-log_{10}2.5=2$ (b) $log_{8}60-log_{8}3-log_{8}5=log_{8}60-(log_{8}3+log_{8}5)$ $=log_{8}60-log_{8}(3\times5)$ $=log_{8}60-log_{8}15$ Hence, $log_{8}60-log_{8}3-log_{8}5=\frac{2}{3}$
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