Question

If (9^n X 3^2 X 3^n -27^n)/ 3^3m X 2^3 = 1/729, prove that m-n is 2.

Answer 1

$\Large \textbf{\underline{\underline{Question:}}}$

$\textsf{If}\ \ \ \dfrac{9^n\times 3^2\times 3^n-27^n}{3^{3m}\times 2^3}=\dfrac{1}{729},\\ \\ \\ \textsf{then prove that \ m-n=2 \ .}$

$\cline{1-}$

$\begin{aligned}&\dfrac{9^n\times 3^2\times 3^n-27^n}{3^{3m}\times 2^3}=\dfrac{1}{729}\\ \\ \Longrightarrow\ \ &\frac{(3^2)^n\times 3^2\times 3^n-(3^3)^n}{3^{3m}\times 2^3}=\frac{1}{729}\\ \\ \Longrightarrow\ \ &\frac{3^{2n}\times 3^2\times 3^n-3^{3n}}{3^{3m}\times 2^3}=\frac{1}{729}\\ \\ \Longrightarrow\ \ &\frac{3^{2n+2+n}-3^{3n}}{3^{3m}\times 2^3}=\frac{1}{729}\\ \\ \Longrightarrow\ \ &\frac{3^{3n+2}-3^{3n}}{3^{3m}\times 2^3}=\frac{1}{729}\end{aligned}$

$\begin{aligned}\Longrightarrow\ \ &\frac{3^{3n}\times 3^2-3^{3n}}{3^{3m}\times 2^3}=\frac{1}{729}\\ \\ \Longrightarrow\ \ &\frac{3^{3n}(3^2-1)}{3^{3m}\times 2^3}=\frac{1}{729}\\ \\ \Longrightarrow\ \ &\frac{3^{3n}(9-1)}{3^{3m}\times 2^3}=\frac{1}{729}\\ \\ \Longrightarrow\ \ &\frac{3^{3n}\times 8}{3^{3m}\times 8}=\frac{1}{729}\\ \\ \Longrightarrow\ \ &\frac{3^{3n}}{3^{3m}}=\frac{1}{3^6}\\ \\ \Longrightarrow\ \ &\left(\frac{3^n}{3^m}\right)^3=3^{-6}\\ \\ \Longrightarrow\ \ &(3^{n-m})^3=3^{-6}\end{aligned}$

$\begin{aligned}\Longrightarrow\ \ &((3^{n-m})^3)^{-\frac{1}{3}}=(3^{-6})^{-\frac{1}{3}}\\ \\ \Longrightarrow\ \ &(3^{n-m})^{-1}=3^2\\ \\ \Longrightarrow\ \ &3^{m-n}=3^2\\ \\ \Longrightarrow\ \ &\large \bold{m-n=2}\end{aligned}$

$\huge\boxed{\ \ \ \ \ \ \ \ \ \ \ \ \textsc{\underline{\underline{Hence Proved!!!}}}\ \ \ \ \ \ \ \ \ \ \ \ }$