Question

plz write correct answer i will mark as brainliest ​

Answer 1

♕ $\red{\Huge{\underline{\underline{ \rm{Solution }}}}}$♕

$\tt{x = \dfrac{ {27}^{3n - 1} . \: {243}^{ \frac{ - 4n}{5} }}{ {9}^{n + 1}. \: {3}^{3n - 5} } }$

We know

◔ 27 = 3 × 3 × 3 = 3³

◔ 243 = 3 × 3 × 3 × 3 ×3 = 3⁵

◔ 9 = 3 × 3 = 3²

Put the values, we have:

${ \tt{x = \dfrac{ ({ {3}^{3}) }^{3n - 1}. \: {( {3}^{5}) }^{ \frac{ - 4n}{5} } }{ {({3}^{2}) }^{n + 1} . \: {3}^{3n - 5} } }}$

$\tt = \dfrac{ {3}^{9n - 3} \times \: {3}^{ - 4n} }{ {3}^{2n + 2} \times \: {3}^{3n - 5} }$

When the bases are same with multiplication sign between, then powers or exponents are added, so we have:

$\tt = \dfrac{ {3}^{9n - 3 + ( - 4)} }{ {3}^{2n + 2 + (3n - 5)} }$

$\tt = \dfrac{ {3}^{9n - 3 - 4n} }{ {3}^{2n + 2 + 3n - 5} }$

$\tt = \dfrac{ {3}^{5n - 3} }{ {3}^{5n - 3} }$

On further solving and cancelling it, we have:

$\tt = 1$

Answer :-

$\green{ \therefore{ \underline{ \boxed{ \tt{x = 1}}}}}$

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Answer 2

$\sf{\huge{\underline{\boxed{\green{\bold{Solution}}}}}}$

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: $\sf\implies \: {\bold{ \dfrac{ 27^{3n-1} . 243^{\frac{- 4n}{5}}}{ 9^{n+1} . 3^{3n-5}}}}$

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• We will find the factors of each term .

27 = 3 x 3 x 3

243 = 3 x 3 x 3 x 3 x 3

9 = 3 x 3

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: $\sf\implies \: {\bold{ \dfrac{ 3^{3( 3n-1 ) } . 3^{5 \times \frac{- 4n}{5}}}{ 3^{2 ( n+1 ) } . 3^{3n-5}}}}$

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: $\sf\implies \: {\bold{ \dfrac{ 3^{9n-3} . 3^{{4n}}}{ 3^{2n+2} . 3^{3n-5}}}}$

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• When the bases are same in multiplication powers will add

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: $\sf\implies \: {\bold{ \dfrac{ 3^{ 9n -3 - 4n}}{3^{2n + 2 + 3n - 5}}}}$

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: $\sf\implies \: {\bold{\dfrac{ 3^{5n - 3 }} { 3^{5n - 3 }}}}$

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• When the bases are same in division the powers will subtract

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:$\sf\implies \: {\bold{ 3^{ 5n - 3 - 5n + 3 }}}$

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:$\sf\implies \: {\bold{ 3^{ 0 }}}$

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• $\sf{\bold{ a^0 = 1 }}$

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:$\sf\implies \: {\bold{ 1 }}$

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$\sf{{\underline{\boxed{\blue{{\bold{ \dfrac{ 27^{3n-1} . 243^{\frac{- 4n}{5}}}{ 9^{n+1} . 3^{3n-5}}}} = 0 }}}}}$

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