Question

Se log 9 na base 1/3=a, então log a² na base 16 é:​

Answer 1

Resposta:

$log_{ \frac{1}{3} }9 = a \\ (\frac{1}{3}) {}^{a} = 9 \\ ({3}^{ - 1}) {}^{a} = {3}^{2} \\ - a = 2 \: \: \: \: \: \: \: .( - 1) \\ a = - 2 \\ {a}^{2} = ( - 2) {}^{2} \\ {a}^{2} = 4$