Question

Solve the following system of equations.
${\log }_{5}x+3^{\log y}=7,x^y={625}^3.$

Answer 1

${\log }_{5}x+3^{{\log }_{3}y}=7$

${\log }_{5}x+y=7$

${\log }_{5}x=(7-y)\to 1$

$x^y={625}^3$

Taking ${\log }_5$ both sides

$\left({\log }_{5}x\right)y=3{\log }_5\left(625\right)$

$y\left({\log }_{5}x\right)=\frac{12}{y}\to 2$

From $1$ and $2$

$\frac{12}{y}=(7-y)$

$12=7y-y^2\Rightarrow y^2-7y+12=0$

$(y-3)(y-4)=0$

$y=3,4$

From $2$

${\log }_{5}x=\frac{12}{3}=4;{\log }_{5}x=\frac{12}{4}=3$

$x=5^4;x=5^3$

$(x,y):(5^4,3),(5^3,4)$