Question

log 3 na base x = 4 e log 5 na base y= 6, qual o valor de (xy) elevado na 12???

Answer 1

$log_x3 = 4 =\ \textgreater \ x^{4} = 3 =\ \textgreater \ x = 3^{ \frac{1}{4}} \\ log_y5 = 6 =\ \textgreater \ y^{6} = 5 =\ \textgreater \ y = 5^{ \frac{1}{6} } \\ (x.y)^{12} =( 3^{ \frac{1}{4}}. 5^{ \frac{1}{6}} )^{12}} \\ 3^{ \frac{12}{4}} .5^{ \frac{12}{6} } \\ 3^{3} . 5^{2} = 27.25 = 675$

Espero ter ajudado.