Bunuel
\(9^{3}(\frac{2^{8} + 2^{9}}{2})=\)
(A) \(2^{7}3^{6}\)
(B) \(2^{6}3^{7}\)
(C) \(6^{7}\)
(D) \(2^{8}3^{7}\)
(E) \(12^{4}\)
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:When working with exponents, you should try to find common (typically prime) bases and multiply. First to multiply, you can factor a 2^8 out of each additive term within the parentheses to get: \(9^{3} 2^{8} (\frac{(1+2)}{2})\). Then, perform the addition within the parentheses to get: \(9^{3}2^{8} (\frac{3}{2})\).
You can break down \(9^{3}\) to be \((3^{2})^3\), which is \(3^{6}\), and then you have common prime terms for each exponent: \(3^{6}2^{8} (\frac{3}{2})\).
Multiplying out the equation \(3^{6}2^{8} (\frac{3}{2})\), you get: \(3^7*3=3^8\) and \(\frac{2^{8}}{2} =2^7\), which can be simplified to \(3^7*2^7\).
This can be expressed as \((3*2)^{7} = 6^{7}\), and
the correct answer is C.
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