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: 45



Work Step by Step

We are given that $log_{10}2=.3010$ and $log_{10}9=.9542$. 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, $log_{10}3=log_{10}9^{\frac{1}{2}}=\frac{1}{2}log_{10}9=\frac{1}{2}(.9542)=.4771$.
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