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: 8


(a) $\log_{10}(10)^{\frac{1}{2}}=\frac{1}{2}$ (b) $\log_{8}320-log_{8} 5=\log_{8}(2^{6})$

Work Step by Step

(a) $\log_{10}\sqrt 10= \log_{10}(10)^{\frac{1}{2}}$ Sinnce, $logx^{y}=xlogy$ $\log_{10}(10)^{\frac{1}{2}}=\frac{1}{2}\log_{10}(10)$ Use logarithmic base formula $\log_{b}x=\frac{log x}{log b}$ $\log_{10}(10)^{\frac{1}{2}}=\frac{1}{2}\frac{log 10}{log 10}$ Hence, $\log_{10}(10)^{\frac{1}{2}}=\frac{1}{2}$ (b) Since, $\log_{b}x=\frac{log x}{log b}$ $\log_{8}320-log_{8} 5=\log_{8}(2^{6}.5)-log_{8} 5$ $\log_{8}320-log_{8} 5=\log_{8}(2^{6})+\log_{8}5-log_{8} 5$ Hence, $\log_{8}320-log_{8} 5=\log_{8}(2^{6})$
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