Question

9^nx3²x[3-n/2]-²-(27)^n/3³m×2³=1/27
If
then prove that m-n=1​

Answer 1

$\large\underline{\sf{Given- }}$

$\: \: \: \: \: \: \: \: \bull \sf \: \dfrac{ {9}^{n} \times {3}^{2} \times {( {3}^{ - \frac{n}{2} }) }^{ - 2} - {27}^{n} }{ {3}^{3m} \times {2}^{3}} = \dfrac{1}{27}$

$\large\underline{\sf{To\:prove - }}$

$\: \: \: \: \: \: \: \: \bull \sf \: m - n = 1$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

Laws of exponents :-

$\begin{gathered}(1)\:{\underline{\boxed{\bf{\blue{a^m\times{a^n}\:=\:a^{m\:+\:n}\:}}}}} \\ \end{gathered}$

$\begin{gathered}(2)\:{\underline{\boxed{\bf{\purple{\dfrac{a^m}{a^n}\:=\:a^{m\:-\:n}\:}}}}} \\ \end{gathered}$

$\begin{gathered}(3)\:{\underline{\boxed{\bf{\orange{\dfrac{1}{x^n}\:=\:x^{-n}\:}}}}} \\ \end{gathered}$

$\begin{gathered}(4)\:{\underline{\boxed{\bf{\color{peru}{(a^m)^n\:=\:a^{m\times{n}}\:}}}}} \\ \end{gathered}$

$\begin{gathered}(5)\:{\underline{\boxed{\bf{\red{ {x}^{0} = 1}}}}} \\ \end{gathered}$

$\begin{gathered}(6)\:{\underline{\boxed{\bf{\green{ {a}^{m} = {a}^{n} \implies \: m \: = \: n}}}}} \\ \end{gathered}$

$\large\underline{\bf{Solution-}}$

↝ Given that

$\rm :\longmapsto\:\sf \: \dfrac{ {9}^{n} \times {3}^{2} \times {( {3}^{ - \frac{n}{2} }) }^{ - 2} - {27}^{n} }{ {3}^{3m} \times {2}^{3}} = \dfrac{1}{27}$

$\rm :\longmapsto\:\sf \: \dfrac{ {(3 \times 3)}^{n} \times {3}^{2} \times {( {3}^{n}) } - {(3 \times 3 \times 3)}^{n} }{ {3}^{3m} \times {2}^{3}} = \dfrac{1}{3 \times 3 \times 3}$

$\rm :\longmapsto\:\sf \: \dfrac{ {3}^{2n} \times {3}^{2} \times {( {3}^{ n}) }- {3}^{3n} }{ {3}^{3m} \times {2}^{3}} = \dfrac{1}{ {3}^{3} }$

$\rm :\longmapsto\:\dfrac{( {3}^{2n} \times {3}^{n}) \times {3}^{2} - {3}^{3n} }{ {3}^{3m} \times 8 } = {3}^{ - 3}$

$\rm :\longmapsto\:\dfrac{( {3}^{2n + n}) \times {3}^{2} - {3}^{3n} }{ {3}^{3m} \times 8 } = {3}^{ - 3}$

$\rm :\longmapsto\:\dfrac{( {3}^{3n}) \times {3}^{2} - {3}^{3n} }{ {3}^{3m} \times 8 } = {3}^{ - 3}$

$\rm :\longmapsto\:\dfrac{{3}^{3n}({3}^{2} - 1)}{ {3}^{3m} \times 8 } = {3}^{ - 3}$

$\rm :\longmapsto\:\dfrac{ {3}^{3n}(9 - 1) }{ {3}^{3m} \times 8 } = {3}^{ - 3}$

$\rm :\longmapsto\:\dfrac{ {3}^{3n} \times \cancel8 }{ {3}^{3m} \times \cancel8 } = {3}^{ - 3}$

$\rm :\longmapsto\:\dfrac{ {3}^{3n} }{ {3}^{3m} } = {3}^{ - 3}$

$\rm :\longmapsto\: {3}^{(3n - 3m)} = {3}^{ - 3}$

$\rm :\implies\:3n - 3m = - 3$

$\rm :\longmapsto\: - 3(m - n) = - 3$

$\bf\implies \:m - n = 1$

${\boxed{\boxed{\bf{Hence, Proved}}}}$