Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 12 - Section 12.7 - Common Logarithms, Natural Logarithms, and Change of Base - Exercise Set: 18

Answer

-1

Work Step by Step

Based on the definition of the common logarithm, we know that $log(x)=log_{10}x$. Furthermore, recall that $log_{b}x=y$ is equivalent to the statement $b^{y}=x$ (where $x\gt0$, $y$ is a real number, and $b\gt0$ and $b\ne1$). Therefore, $log(\frac{1}{10})=log_{10}\frac{1}{10}=-1$, because $10^{-1}=\frac{1}{10^{1}}=\frac{1}{10}$.
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