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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.

The correct answer is A.

Explanation:

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 \(2^{12}\).


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