Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 9 - Section 9.4 - Properties of Logarithms - 9.4 Exercises: 42

Answer

2.2094

Work Step by Step

We are given that $log_{10}2=.3010$ and $log_{10}9=.9542$. We know that $log_{b}xy=log_{b}x+log_{b}y$ (where $x$, $y$, and $b$ are positive real numbers and $b\ne1$). Therefore, $log_{10}162=log_{10}(81\times2)=log_{10}81+log_{10}2$. We know that $log_{b}x^{r}=rlog_{b}x$ (where $x$ and $b$ are positive real numbers, $b\ne1$, and $r$ is a real number). Therefore, $log_{10}81+log_{10}2=log_{10}9^{2}+log_{10}2=2log_{10}9+log_{10}2=2(.9542)+.3010=1.9084+.3010=2.2094$.
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