Question

9n×3²×(3-n/2)-²-(27)n/3³m×2³=1/27 prove m-n=1​

Answer 1

Appropriate Question :-

$\sf \: If \: \dfrac{ {9}^{n} \times{3}^{2} {\bigg( {3}^{ - \frac{n}{2} } \bigg) }^{ - 2} - {27}^{n}}{ {3}^{3m} \times {2}^{3} } = \dfrac{1}{27}, \: prove \: that \: m - n = 1$

Basic Concept Used :-

Law of exponents :-

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

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

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

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

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

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

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

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

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

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

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

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

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

$\rm :\longmapsto\:\dfrac{ {3}^{3n}}{ {3}^{3m}} = \dfrac{1}{ {3}^{3} }$

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

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

$\rm :\longmapsto\:n - m = - 1$

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

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