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# If $$3x - y = 12$$, what is the value of $$\frac{8^x}{2^y}$$?

### Question

If $$3x - y = 12$$, what is the value of $$\frac{8^x}{2^y}$$?

### Options

A)
$$2^{12}$$
B)
$$4^4$$
C)
$$8^2$$
D)
The value cannot be determined from the information given.

Choice A is correct. One approach is to express $$\frac{8^x}{2^y}$$ so that the numerator and denominator are expressed with the same base. Since 2 and 8 are both powers of 2, substituting $$2^3$$ for 8 in the numerator of $$\frac{8^x}{2^y}$$ gives $$\frac{(2^3)^x}{2^y}$$, which can be rewritten as $$\frac{2^{3x}}{2^y}$$. Since the numerator and denominator of $$\frac{2^{3x}}{2^y}$$ have a common base, this expression can be rewritten as $$2^{3x-2y}$$. It is given that 3x-y = 12, so one can substitute 12 for the exponent, 3x - y, given that the expression $$\frac{8^x}{2^y}$$ is equal to $$2^{12}$$.