Question

Calcule o log₂64 + log₃81​

Answer 1

$log_{2}(64) =log_{{2}^{1}}({2}^{6})=\frac{6}{1}=6$

$log_{3}(81) =log_{{3}^{1}}({3}^{4}) =\frac{4}{1}=4$

$\boxed{\boxed{log_{2}(64)+log_{3}(81)=6+4=10}}$

Answer 2

Resposta:

log₂ 64 + log₃ 81​

log₂ 2⁶ + log₃ 3⁴

=6* log₂ 2 +4*log₃ 3 =6+4=10

Propriedades utilizadas

log(a) a = 1    ...sendo a>0   ..(a)  é a base

log b^c =c* log b