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Mathematics

If 9n×32×3n−(27)n33m×23=3−3\dfrac{9^n \times 3^2 \times 3^n - (27)^n}{3^{3m} \times 2^3} = 3^{-3}, prove that (m - n) = 1.

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Answer

Given,

9n×32×3n−(27)n33m×23=3−3\dfrac{9^n \times 3^2 \times 3^n - (27)^n}{3^{3m} \times 2^3}= 3^{-3}

Solving:

⇒9n×32×3n−(27)n33m×23=3−3⇒[(3)2]n×32×3n−[(3)3]n33m×8=3−3⇒(3)2n×32×3n−(3)3n33m×8=3−3⇒(3)2n+2+n−(3)3n33m×8=3−3⇒(3)3n+2−(3)3n33m×8=3−3⇒33n.32−33n33m×8=3−3⇒33n[(32)−1]33m×8=3−3⇒33n[9−1]33m×8=3−3⇒33n×833m×8=3−3⇒33n−3m=3−3⇒33(n−m)=3−3⇒3−3(m−n)=3−3\Rightarrow \dfrac{9^n \times 3^2 \times 3^n - (27)^n}{3^{3m} \times 2^3} = 3^{-3} \\[1em] \Rightarrow \dfrac{[(3)^2]^n \times 3^2 \times 3^n - [(3)^3]^n}{3^{3m} \times 8} = 3^{-3} \\[1em] \Rightarrow \dfrac{(3)^{2n} \times 3^2 \times 3^n - (3)^{3n}}{3^{3m} \times 8} = 3^{-3} \\[1em] \Rightarrow \dfrac{(3)^{2n + 2 + n} - (3)^{3n}}{3^{3m} \times 8} = 3^{-3} \\[1em] \Rightarrow \dfrac{(3)^{3n + 2} - (3)^{3n}}{3^{3m} \times 8} = 3^{-3} \\[1em] \Rightarrow \dfrac{3^{3n}.3^2 - 3^{3n}}{3^{3m} \times 8} = 3^{-3}\\[1em] \Rightarrow \dfrac{3^{3n}[(3^2) - 1]}{3^{3m} \times 8} = 3^{-3} \\[1em] \Rightarrow \dfrac{3^{3n}[9 - 1]}{3^{3m} \times 8} = 3^{-3} \\[1em] \Rightarrow \dfrac{3^{3n} \times 8}{3^{3m} \times 8} = 3^{-3} \\[1em] \Rightarrow 3^{3n - 3m} = 3^{-3} \\[1em] \Rightarrow 3^{3(n - m)} = 3^{-3} \\[1em] \Rightarrow 3^{-3(m - n)} = 3^{-3} \\[1em]

Equating the exponents,

⇒ -3(m - n) = -3

⇒ (m - n) = −3−3\dfrac{-3}{-3} = 1.

Hence proved, m - n = 1.

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