Look For The Bases

Relevant wiki: Rules of Exponents - Algebraic

$\Rightarrow \dfrac{8^x}{2^y}$

$\dfrac{2^{3x}}{2^y}$

$2^{3x-y}$

$2^{12}=\boxed{4096}$

$3x-y=12 \implies y=3x-12$

$\dfrac{8^x}{2^y}\\ =\dfrac{(2^3)^x}{2^{3x-12}}\\ =2^{3x-(3x-12)}\\ =2^{12}\\ =\boxed{4096}$

One approach is to show

$\dfrac{8^{x}}{2^{y}}$

so that the numerator and denominator are expressed with the same base. Since 2 and 8 are both powers of 2, changing 8 into $2^{3}$ gives us

$\dfrac{(2^{3})^{x}}{2^{y}}$

which we can change to

$\dfrac{8^{3}{x}}{2^{y}}$

Since it now shares a common base we can change it to $2^{3x - y}$ , which we can change to $2^{12}$ , based on 3 $x$ - $y$ = 12.

$2^{12}$ = $\boxed{4096}$

y=3x-12--> 8=2^(3x)-->[2^(3x)]/[2^(3x-12)]=2^(12)=4096. I hate formatting.

3x-y=12 => x=12+y/3 [(2^3)^(12+y/3) ]/2^y ==> 2^(12+y)/2^y , ==> 2^12+y-y ==> 2^12=4096