## Intermediate Algebra (6th Edition)

$log_{6}\frac{2}{3}\approx-.2263$
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, $log_{6}\frac{2}{3}=\frac{log_{10}\frac{2}{3}}{log_{10}6}=\frac{-.1761}{.7782}\approx-.2263$.