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#