# How do you evaluate log_(1/3) (1/9)?

##### 2 Answers
Sep 6, 2016

Rewrite ${\log}_{\frac{1}{3}} \left(\frac{1}{9}\right) = x$ as ${\left(\frac{1}{3}\right)}^{x} = \left(\frac{1}{9}\right)$.

#### Explanation:

Rewrite ${\log}_{\frac{1}{3}} \left(\frac{1}{9}\right) = x$ as ${\left(\frac{1}{3}\right)}^{x} = \left(\frac{1}{9}\right)$.

Remember, the answer to a log is an exponent, so you are looking for the exponent that makes $\frac{1}{3}$ turn into $\frac{1}{9}$.

$\frac{1}{3}$ is both the base of the log and the base of the exponent.

${\left(\frac{1}{3}\right)}^{x} = \left(\frac{1}{9}\right)$.

$x = 2$ because ${\left(\frac{1}{3}\right)}^{2} = \frac{1}{9}$.

Sep 6, 2016

${\log}_{\frac{1}{3}} \left(\frac{1}{9}\right) = 2$

#### Explanation:

In this log form of ${\log}_{\frac{1}{3}} \left(\frac{1}{9}\right)$

The question being asked is
"what power or index of $\frac{1}{3}$ will give $\frac{1}{9}$"?

This one can be done by inspection.

The answer is clearly 2!

Note that ${\left(\frac{1}{3}\right)}^{2} = \frac{1}{9}$

${\log}_{\frac{1}{3}} \left(\frac{1}{9}\right) = 2$

Else you have to use the change of base law:

$\frac{\log \left(\frac{1}{9}\right)}{\log \left(\frac{1}{3}\right)} = 2$