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

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Answer 1 · 2016-09-06T11:31:38

Rewrite $log_(1/3) (1/9) =x$ as $(1/3)^x = (1/9)$ .

Rewrite $log_(1/3) (1/9) =x$ as $(1/3)^x = (1/9)$ .

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

$1/3$ is both the base of the log and the base of the exponent.

$(1/3)^x = (1/9)$ .

$x =2$ because $(1/3)^2 = 1/9$ .

Answer 2 · 2016-09-06T11:52:04

$log_(1/3) (1/9) = 2$

In this log form of $log_(1/3) (1/9)$

The question being asked is
"what power or index of $1/3$ will give $1/9$ "?

This one can be done by inspection.

The answer is clearly 2!

Note that $(1/3)^2 = 1/9$

$log_(1/3) (1/9) = 2$

Else you have to use the change of base law:

$(log(1/9))/(log(1/3)) = 2$

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