How do you find the exact value of ${log}_{3} 81^{- 0.2}$ $log_3 81^-0.2$ ?

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Answer 1 · 2016-12-29T07:27:57

$- 0.8$ $-0.8$

Since ${log a}^{b} = b log a$ $loga^b=bloga$ ,

the given expression becomes:

$- 0.2 {log}_{3} 81$ $-0.2log_3 81$

$- 0.2 {log}_{3} 3^{4}$ $-0.2log_3 3^4$

$- 0.2 \cdot 4$ $-0.2*4$

$- 0.8$ $-0.8$

How do you find the exact value of log_3 81^-0.2log381−0.2log_3 81^-0.2?