Question

resolver o seguinte sistema de equação:

4 elevado a x/2 +y = 32
log de x na base 2 + log y na base 1/2 = -1

Answer 1

$\left \{ {{4^{x^{2}+y}=32} \atop {\log_2x+\log_ \frac{1}{2}y=-1 }} \right.$

$\log_2x+\log_ \frac{1}{2}y=-1$
$\log_2x+\log_ {2^{-1}}y=-1$
$\log_2x-\log_2y=-1$
$\log_2 \frac{x}{y} =-1$
$\frac{x}{y} = 2^{-1}$
$\frac{x}{y} = \frac{1}{2}$
$y=2x$

$4^{ \frac{x}{2}+y}=32$
$4^{\frac{x}{2}+2x}=32$
$2^{2}^{ (\frac{x}{2}+2x)}= 2^{5}$
$2( \frac{x}{2}+2x)=5$
$x+4x=5$
$5x=5$
$x=1$

$y=2x$
$y=2.1$
$y=2$