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: $log_81(1/3) = -1/4$

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