How do you evaluate $log_(1/3) (1/81)$ ?

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Answer 1 · 2016-12-16T02:06:40

$4$

$1/81$ can be written as $(81)^-1$ :

$= log_(1/3) (81^-1)$

Use the change of base rule $log_a n = logn/loga$ :

$= log81^-1/log(1/3)$

$=log(3^4)^-1/log(3^-1)$

$= log(3^-4)/log3^-1$

Use the exponent rule that $loga^n = nloga$ :

$=(-4log3)/(-1log3)$

$=4$

Hopefully this helps!

How do you evaluate log_(1/3) (1/81)log_(1/3) (1/81)?