Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 9 - Section 9.7 - Common Logarithms, Natural Logarithms, and Change of Base - Exercise Set - Page 585: 33

Answer

$x=10^{1.3}\approx19.9526$

Work Step by Step

We are given the equation $\log(x)=1.3$. To solve for x, remember that the base of a common logarithm is understood to be 10. Therefore, $log(x)=log_{10}x=1.3$ If $b\gt0$ and $b\ne1$, then $y=log_{b}x$ means $x=b^{y}$ for every $x\gt0$ and every real number $y$. Therefore, $x=10^{1.3}\approx19.9526$.
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