How do you find the exact value of $log_3 81^-0.2$ ?

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Answer 1 · 2016-12-29T07:27:57

$-0.8$

Since $loga^b=bloga$ ,

the given expression becomes:

$-0.2log_3 81$

$-0.2log_3 3^4$

$-0.2*4$

$-0.8$

How do you find the exact value of log_3 81^-0.2log_3 81^-0.2?