Bunuel wrote:
If n is a number such that \((−8)^{(2n)} = 2^{(8+2n)}\), then n =
(A) 1/2
(B) 3/2
(C) 2
(D) 4
(E) 5
\((−8)^{(2n)} = 2^{(8+2n)}\)
Any number with negative raised to even number exponent will always be positive. Therefore\((−8)^{(2n)}\) can be written as\((8)^{(2n)}\)
\((8)^{(2n)} = 2^{(8+2n)}\)
\(2^3^{(2n)} = 2^{(8+2n)}\)
\(2^{6n} = 2^{(8+2n)}\)
Therefore;
6n = 8 + 2n
6n - 2n = 8
4n
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