Intermediate Algebra (12th Edition)

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

Chapter 9 - Section 9.5 - Common and Natural Logarithms - 9.5 Exercises: 12

Answer

-0.7592

Work Step by Step

We know that logarithms to base 10 are common logarithms, and $log_{10}x$ is equivalent to $log(x)$. Therefore, $log(.1741)=log_{10}.1741$. We know that for all positive numbers $a$ (where $a\ne1$), and all positive numbers $x$, $y=log_{a}x$ means the same as $x=a^{y}$. Therefore, $log_{10}.1741\approx-0.7592$, because $10^{-0.7592}\approx.1741$.
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