Question

Solve the following system of equations.
$3^x.2^y=576,{\log }_{\sqrt{2}}(y-x)=4.$

Answer 1

${\log }_{\sqrt{2}}(y-x)=4$

$(y-x)=(\sqrt{2}{)}^4$

$(y-x)=4$

$y=4+x$ (equation$1$)

$3^x\cdot 2^y=576$

From $1$,

$\Rightarrow 3^x\cdot 2^{4+x}=576$

$\Rightarrow 16\cdot 6^x=576$

$\Rightarrow 6^x=36$

$\Rightarrow x=2$

From $1$,

$y=6$

$(x,y):(2,6)$