Question

Solve the following system of equations.
$y-{\log }_{3}x=1,x^y=3^{12}$

Answer 1

$y-{\log }_3\left(x\right)=1x^y=3^{12}$

$y-{\log }_3\left(x\right)=1\to 1x^y=3^{12}\to 2$

$x^y=3^{12}$

Taking ${\log }_3$ both side we get

$y{\log }_3\left(x\right)=\left(12\right){\log }_{3}3$

$y\left({\log }_{3}x\right)=12$

$\left({\log }_{3}x\right)=\left(\frac{12}{y}\right)\to 3$

From $1$ and $3$

$y-\frac{12}{y}=1\Rightarrow y^2-y-12=0$

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

$y=-3,4$

From $2$ $x^4=3^{12}$

$x^{-3}=3^{12}{x}^4=3^{12}$

$\left(x\right)=3^{12/-3}x=3^{12/4}$

$x=3^{-4}x=3^3$

$(x,y):(3^{-4},-3)(3^3,4)$