## Intermediate Algebra (6th Edition)

Published by Pearson

# Chapter 9 - Section 9.7 - Common Logarithms, Natural Logarithms, and Change of Base - Exercise Set: 52

#### Answer

$log_{\frac{1}{3}}2\approx-.6309$

#### Work Step by Step

In order to evaluate the given logarithm, we can use the change of base formula. $log_{b}a=\frac{log_{c}a}{log_{c}b}$ (where a, b, and c are positive real numbers and neither b nor c is 1) For simplicity, we will use base 10, which is the base of the common logarithm. Therefore, $log_{\frac{1}{3}}2=\frac{log_{10}2}{log_{10}\frac{1}{3}}=\frac{.3010}{-.4771}\approx-.6309$.

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