Решите систему уравнений 16^x=64 27^x=81^(y-1)

Решите задачу:

$16^x=64\\\\(4^2)^x=4^3\\\\4^{2x}=4^3\iff2x=3\ \ \ /:2\Rightarrow x=\frac{3}{2}\\-------------------\\\\27^x=81^{y-1}\\\\27^\frac{3}{2}=81^{y-1}\\\\\left(3^3\right)^\frac{3}{2}=\left(3^4\right)^{y-1}\\\\3^\frac{9}{2}=3^{4(y-1)}\iff4(y-1)=\frac{9}{2}\ \ \ \ /:4\\\\y-1=\frac{9}{8}\\\\y=\frac{9}{8}+1\\\\y=\frac{9}{8}+\frac{8}{8}\\\\y=\frac{17}{8}\\\\y=2\frac{1}{8}$