Question

EEAR) resolvendo o sistema obtemos:
$log_{2}(x) + log_{4}(y) = 4 \\ xy = 8$
x=?
y=?
​

Answer 1

Resposta:

$log_2(x)+log_2(y)^2=4$

$\frac{2log_2(x)}{2}+\frac{4log_2(y)}{2}=4$

$\frac{6log_2(x*y)}{2}=4$

$log_2(x*y)=\frac{4*2}{6}$

$x*y=\sqrt[3]{16}$

$\left \{ {{x*y=\sqrt[3]{16} } \atop {x+2y=16 }} \right.$

$\left \{ {{x*\frac{16-x}{2} =\sqrt[3]{16} } \atop {y=\frac{16-x}{2} }} \right.$

$\left \{ {-x^2+16x-2\sqrt[3]{16}=0 } } \atop {y=\frac{16-x}{2} }} \right.$

$\left \{ {{x=0,32143791} \atop {y=\frac{16-0,32143791}{2} }} \right.$

$\left \{ {{x=0,32143791} \atop {y=7,839281045 }} \right.$