Question

Solve the following system of equations.
${\log }_{4}x-{\log }_{2}y=0.x^2-2y^2=8.$

Answer 1

${\log }_{4}x-{\log }_{2}y=0$

$\frac{1}{2}{\log }_{2}x-{\log }_{2}y=0$

${\log }_{2}x=2{\log }_{2}y$

$x=y^2\to 1$

$x^2-2y^2=8\to 2$

From $1$ and $2$

$x^2-2x-8=0$

$(x-4)(x+2)=0$

$x=4,-2$

From $1$ ; $x=y^2$

$4=y^2;-2=y^2$

$\pm 2=y;\pm \sqrt{2}i=y$ (non real solution)

$(x,y):(4,2)(4,-2)$