If 9n.32.3n−(27)n33m.23=127\dfrac{9^n.3^2.3^n - (27)^n}{3^{3m}.2^3} = \dfrac{1}{27}33m.239n.32.3n−(27)n=271, prove that m = 1 + n.
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Given,
⇒9n.32.3n−(27)n33m.23=127⇒(32)n.32.3n−(33)n33m.8=133⇒32n+n.32−33n33m.8=133⇒33n.9−33n33m.8=133⇒33n(9−1)33m.8=133⇒33n.833m.8=3−3⇒33n33m=3−3⇒33n=33m.3−3⇒33n=33m+(−3)⇒3n=3m−3⇒3n=3(m−1)⇒n=m−1⇒m=1+n.\Rightarrow \dfrac{9^n.3^2.3^n - (27)^n}{3^{3m}.2^3} = \dfrac{1}{27} \\[1em] \Rightarrow \dfrac{(3^2)^n.3^2.3^n - (3^3)^n}{3^{3m}.8} = \dfrac{1}{3^3} \\[1em] \Rightarrow \dfrac{3^{2n + n}.3^2 - 3^{3n}}{3^{3m}.8} = \dfrac{1}{3^3} \\[1em] \Rightarrow \dfrac{3^{3n}.9 - 3^{3n}}{3^{3m}.8} = \dfrac{1}{3^3} \\[1em] \Rightarrow \dfrac{3^{3n}(9 - 1)}{3^{3m}.8} = \dfrac{1}{3^3} \\[1em] \Rightarrow \dfrac{3^{3n}.8}{3^{3m}.8} = 3^{-3} \\[1em] \Rightarrow \dfrac{3^{3n}}{3^{3m}} = 3^{-3} \\[1em] \Rightarrow 3^{3n} = 3^{3m}.3^{-3} \\[1em] \Rightarrow 3^{3n} = 3^{3m + (-3)} \\[1em] \Rightarrow 3n = 3m - 3 \\[1em] \Rightarrow 3n = 3(m - 1) \\[1em] \Rightarrow n = m - 1 \\[1em] \Rightarrow m = 1 + n.⇒33m.239n.32.3n−(27)n=271⇒33m.8(32)n.32.3n−(33)n=331⇒33m.832n+n.32−33n=331⇒33m.833n.9−33n=331⇒33m.833n(9−1)=331⇒33m.833n.8=3−3⇒33m33n=3−3⇒33n=33m.3−3⇒33n=33m+(−3)⇒3n=3m−3⇒3n=3(m−1)⇒n=m−1⇒m=1+n.
Hence, proved that m = 1 + n.
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