Variables and Exponents

Similar solution as @Michael Wang 's, presented as follows:

$\begin{aligned} \frac {8x}{2^y} & = \frac {2^{3x}}{2^y} \\ & = 2^{\color{#3D99F6}3x-y} & \small \color{#3D99F6} \text{Given that }3x-y = 12 \\ & = 2^{\color{#3D99F6}12} \\ & = \boxed{4096} \end{aligned}$

Solve for $\large{y}$ in terms of $\large{x}$ then substitute:

$\large{3x-y=12}$ $\large{\color{#D61F06}\implies}$ $\large{y=3x-12}$

Substitute:

$\large{\dfrac{8^x}{2^y}=\dfrac{2^{3x}}{2^{3x-12}}=\dfrac{2^{3x}}{2^{3x}2^{-12}}=\dfrac{1}{2^{-12}}=\dfrac{1}{\dfrac{1}{2^{12}}}=\dfrac{1}{1} \times \dfrac{2^{12}}{1}=2^{12}=\boxed{4096}}$

$\frac{8^x}{2^y}=\frac{(2^3)^x}{2^y}=\frac{2^{3x}}{2^y}$ . Since it is given that $3x-y=12$ , it concludes that $3x=12+y$ . Therefore, $\frac{2^{3x}}{2^y}$ becomes $\frac{2^{12+y}}{2^y}$ , or $\frac{2^{12}\cdot 2^y}{2^y}$ , or $2^{12}$ , or $4096$ .