Question

prove that m - n = 1​

Answer 1

Concept;-Focus on LHS and convert it into factors of 3

LHS=

$\frac{ ({9})^{a} \times {3}^{2} {3}^{ { (\frac{ -9}{2} )}^{ -2} } - ( {27})^{a} }{( {3})^{3m} \times {2}^{3} } = \frac{ {3}^{ {2}^{a} } \times {3}^{2 + a} - {3}^{3a} }{ {3}^{3m} \times {2}^{3} } \\ Lhs = \frac{ {3}^{3a + 2} - {3}^{3a} }{ {3}^{3m} \times {2}^{3} } \\ take \: out \: {3}^{3a} \: common \: we \: get \\ Lhs = > \frac{ {3}^{3a} ( {3}^{2} - 1)}{ {3}^{3m} \times {2}^{3} } = \frac{ {3}^{ 3a} \times {2}^{3} }{ {3}^{3m} \times {2}^{3} } = \frac{ {3}^{3a} }{ {3}^{3m} } = \frac{1}{ {3}^{3m - 3a} } \\ but \: it \: is \: equal \: to \: \frac{1}{27} or \: \frac{1}{ {3}^{3} } so \\ \frac{1}{ {3}^{3(m - a)} } = \frac{1}{ {3}^{3 \times 1} } \\ Comparing \: both \: side \: we \: get \: m - a = 1$

Answer 2

☯ Explanation,

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$\bigstar\: \:\red{\bold{\dfrac{9 {}^{n} \times 3 {}^{2} \bigg(3 {}^{ - \dfrac{n}{ 2} } \bigg) {}^{ - 2} - (27 ) {}^{n} }{3 {}^{3m} \times 2 {}^{3} } = \dfrac{1}{27} }}$

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$\displaystyle\longrightarrow \: \bold{ \dfrac{9 {}^{n} \times 3 {}^{2} \bigg(3 {}^{ - \dfrac{n}{ \not2} } \bigg) {}^{ \not{- 2}} - 27 {}^{n} }{3 {}^{3m} \times 2 {}^{3} } = \dfrac{1}{27} }$

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$\displaystyle\longrightarrow \bold{ \dfrac{3 {}^{2n} \times 3 {}^{2} \times 3 {}^{n} - 27 {}^{n} }{3 {}^{3m} \times 2 {}^{3} } = \dfrac{1}{27} }$

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$\displaystyle\longrightarrow \bold{ \dfrac{3 {}^{2n + 2 + n} - 3 {}^{3n} }{3 {}^{3m} \times 2 {}^{3} } = \dfrac{1}{27} }$

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$\displaystyle\longrightarrow \bold{ \dfrac{3 {}^{3n + 2 } - 3 {}^{3n} }{3 {}^{3m} \times 2 {}^{3} } = \dfrac{1}{27} }$

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$\displaystyle\longrightarrow \bold{ \dfrac{3 {}^{3n} \times (3 {}^{2} - 1) }{3 {}^{3m} \times 2 {}^{3} } = \dfrac{1}{27} }$

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$\displaystyle\longrightarrow \bold{ \dfrac{3 {}^{3n} \times 8}{3 {}^{3m} \times 8 } = \dfrac{1}{27} }$

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$\displaystyle\longrightarrow \bold{ \dfrac{3 {}^{3n} \times \not8}{3 {}^{3m} \times \not8 } = \dfrac{1}{27} }$

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$\displaystyle\longrightarrow \bold{ \dfrac{3 {}^{3n} }{3 {}^{3m} } = \dfrac{1}{27} }$

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$\displaystyle\longrightarrow \bold{ \not3 {}^{3n - 3m} = \not3 {}^{ - 3} }$

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$\displaystyle \longrightarrow \bold{3n - 3m = - 3}$

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$\displaystyle \longrightarrow \bold{ \not{ 3}(n- m) = - \not{3}}$

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$\displaystyle \longrightarrow \bold{ \not{ - }(m- n) = \not{ - }1 }$

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$\displaystyle \longrightarrow \boxed{ \green{\bold{m - n = 1 }}} \: \bigstar$

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