Question

〖Log〗_8 8^(-2) Log de 8 na base 8 elevada a potência -2 pf urgente !!

Answer 1

$log_{b}(a^{n})=n*log_{b}(a)$
$log_{n}(n)=1$
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$log_{8}(8^{-2})=(-2)*log_{8}(8)$
$log_{8}(8^{-2})=(-2)*1$
$log_{8}(8^{-2})=-2$