How do you evaluate #log_81 (1/3)#?

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Answer 1 · 2016-08-20T04:15:46

#log_81(1/3) = -1/4#

Let #x =log_81(1/3)#

Then: #81^x = 1/3#

Since #81 =3^4# and #1/3 = 3^(-1) ->#

#3^(4x) = 3^(-1)#

Equating indices: #4x=-1#

#x=-1/4#

Therefore: #log_81(1/3) = -1/4#