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

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

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Answer 1 · 2016-12-16T02:06:40

$4$ $4$

Explanation:

$\frac{1}{81}$ $1/81$ can be written as ${(81)}^{- 1}$ $(81)^-1$ :

$= {log}_{\frac{1}{3}} (81^{- 1})$ $= log_(1/3) (81^-1)$

Use the change of base rule ${log}_{a} n = \frac{log n}{log a}$ $log_a n = logn/loga$ :

$= \frac{{log 81}^{- 1}}{log (\frac{1}{3})}$ $= log81^-1/log(1/3)$

$= \frac{{log (3^{4})}^{- 1}}{log (3^{- 1})}$ $=log(3^4)^-1/log(3^-1)$

$= \frac{log (3^{- 4})}{{log 3}^{- 1}}$ $= log(3^-4)/log3^-1$

Use the exponent rule that ${log a}^{n} = n log a$ $loga^n = nloga$ :

$= \frac{- 4 log 3}{- 1 log 3}$ $=(-4log3)/(-1log3)$

$= 4$ $=4$

Hopefully this helps!

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