How do you evaluate #log_(1/13) (1/169)#?

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How do you evaluate #log_(1/13) (1/169)#?

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Answer 1 · 2017-01-20T11:04:34

#2#

Explanation:

Spotting that #169=13^2# so you need #log_(1/13)(1/13^2)#.
Since #log_a(x^2)=2log_a(x)# and #log_a(a)=1# you get
#log_(1/13)((1/13)^2)=2log_(1/13)(1/13)=2 xx 1 = 2#.

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How do you evaluate #log_(1/13) (1/169)#?

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