Solve ,for x and y, the pair of simultaneous equations
8^x / 2^y = 64
3^4x * (1/9) ^y-1 = 81
$\dfrac{8^x}{2^y}=64$
$\dfrac{2^{3x}}{2^y} = 64$
$2^{3x-y} = 64$
$3x-y = 6$
$3^{4x} \left(\dfrac{1}{9}\right)^{y-1} = 81$
$\dfrac{3^{4x}}{3^{2y-2}} = 81$
$3^{4x - 2y+2} = 81$
$4x - 2y +2 = 4$
$4x - 2y = 2$
So now 2 linear equations
$3x-y = 6$
$4x-2y=2$
Spoiler: