Given that \(\frac{1}{8^{2-3y}}\) = 2\(^{y + 2}\), find y.
\(\frac{1}{8^{2-3y}}\) = 2\(^{y + 2}\)
8\(^{-1(2 - 3y)}\) = 2\(^{y + 2}\)
2\(^{3(-2 + 3y)}\) = 2\(^{y + 2}\)
-6 + 9y = y + 2
9y - y = 2 + 6
8y = 8
y = \(\frac{8}{8}\) = 1
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