If \(\frac{3^{(1-n)}}{9^{-2n}}\) = \(\frac{1}{9}\) find n
\(\frac{3^{(1-n)}}{9^{-2n}}\) =\(\frac{1}{9}\) 3\(^{1-n}\times 3^{-2(-2n)} = 3^{-2}\) 1 - n - 2(-2n) = -2 1 - n + 4n = -2 n = -1
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