How do you evaluate ${log}_{\frac{1}{3}} (\frac{1}{9})$ $log_(1/3) (1/9)$ ?

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Answer 1 · 2016-09-06T11:31:38

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

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

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

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

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

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

Answer 2 · 2016-09-06T11:52:04

${log}_{\frac{1}{3}} (\frac{1}{9}) = 2$ $log_(1/3) (1/9) = 2$

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

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

This one can be done by inspection.

The answer is clearly 2!

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

${log}_{\frac{1}{3}} (\frac{1}{9}) = 2$ $log_(1/3) (1/9) = 2$

Else you have to use the change of base law:

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

How do you evaluate log_(1/3) (1/9)log13(19)log_(1/3) (1/9)?