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 - Page 614: 37

Answer

1.2552

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, $log_{10}18=log_{10}(9\times2)=log_{10}9+log_{10}2=.3010+.9542=1.2552$.
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