## Algebra: A Combined Approach (4th Edition)

Published by Pearson

# Chapter 12 - Section 12.7 - Common Logarithms, Natural Logarithms, and Change of Base - Exercise Set: 39

#### Answer

$\frac{10^{1.1}}{2}$, $6.2946$

#### 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, $log(2x)=log_{10}2x=1.1$ is equivalent to $10^{1.1}=2x$. Divide both sides by 2. $x=\frac{10^{1.1}}{2}\approx6.2946$

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