Hard

One approach is to express

$\frac{8x}{2y}$

so that the numerator and denominator are expressed in the same base. Since 2 and 8 are both powers of 2, substituting 23 for 8 in the numerator of $\frac{8x}{2y}$ gives

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

which can be rewritten

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

Since the numerator and denominator of have a common base, this expression can be rewritten as 2(3x−y). In the question, it states that 3x−y=12, so one can substitute 12 for the exponent, 3x−y, giving that the

$\frac{8^x}{2^y}$ =212