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If \[\LARGE{\color{red}{(9^n)(3^2)(3^{-n/2})^{-3}-(\sqrt{177147})^n} \over 3^{3m}(2)^3}=\frac{1}{27}\] then:
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so let us take \(9^n=3^{2n}\) hence put \(9^n=3^{2n}\)
a) m-n+2=0 b) 6m+11n-6=0 c)6m-11n-6=0 d)m-n-2=0
|dw:1342880857921:dw|
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