How do you evaluate ${log}_{81} (\frac{1}{3})$ $log_81 (1/3)$ ?

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Answer 1 · 2016-08-20T04:15:46

${log}_{81} (\frac{1}{3}) = - \frac{1}{4}$ $log_81(1/3) = -1/4$

Let $x = {log}_{81} (\frac{1}{3})$ $x =log_81(1/3)$

Then: $81^{x} = \frac{1}{3}$ $81^x = 1/3$

Since $81 = 3^{4}$ $81 =3^4$ and $\frac{1}{3} = 3^{- 1} \to$ $1/3 = 3^(-1) ->$

$3^{4 x} = 3^{- 1}$ $3^(4x) = 3^(-1)$

Equating indices: $4 x = - 1$ $4x=-1$

$x = - \frac{1}{4}$ $x=-1/4$

Therefore: ${log}_{81} (\frac{1}{3}) = - \frac{1}{4}$ $log_81(1/3) = -1/4$

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