# How do you evaluate log_27 (1/3)?

Aug 4, 2016

${\log}_{27} \left(\frac{1}{3}\right) = - \frac{1}{3}$

#### Explanation:

By definition $\log$ is the inverse of exponentiation and vice versa for any base $b$. That is:

${\log}_{b} \left({b}^{x}\right) = x$ for all Real values of $x$

${b}^{{\log}_{b} x} = x$ for all $x > 0$

The change of base formula tells us that for any $a , b , c > 0$ with $a \ne 1$ and $c \ne 1$:

${\log}_{a} b = \frac{{\log}_{c} b}{{\log}_{c} a}$

So we find:

${\log}_{27} \left(\frac{1}{3}\right) = \frac{{\log}_{3} \left(\frac{1}{3}\right)}{{\log}_{3} 27} = \frac{{\log}_{3} {3}^{- 1}}{{\log}_{3} {3}^{3}} = - \frac{1}{3}$