How do you evaluate #log_(1/4) (1/4)#?

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Answer 1 · 2016-07-27T19:14:19

#log_(1/4) (1/4) = 1#

For any base #b > 0# with #b != 1# we have:

#log_b b = 1#

#b=1/4# is no exception.

If you like, use the change of base formula:

#log_a b = (log_c b)/(log_c a)#

So we find:

#log_(1/4) (1/4) = (log (1/4))/(log (1/4)) = 1#