Ans : 27....................... \frac{(243 {)}^(2n/5) . {3}^(2n + 1) }{ {9}^(n + 1). {3}^(2(n - 2)) } \\ please give step by step explanation... ​

Question

Ans : 27.......................


`\frac{(243 {)}^(2n/5) . {3}^(2n + 1) }{ {9}^(n + 1). {3}^(2(n - 2)) } \\`

please give step by step explanation...
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Answer 1

`{ \qquad\qquad\huge\underline{{\sf Answer}}}`

Here we go ~

`\qquad \sf  \dashrightarrow \: \frac{((243 {)}^(2n/5) ) \sdot( {3}^(2n + 1)) }{ ({9}^(n + 1)) \sdot( {3}^(2(n - 2)) )}`

`\qquad \sf  \dashrightarrow \: \frac{(3{)}^(5(2n/5)) ) \sdot( {3}^(2n + 1)) }{ ({3 }^(2(n + 1))) \sdot( {3}^(2(n - 2)) )}`

`\qquad \sf  \dashrightarrow \: \frac{(3{)}^(2n) \sdot( {3}^(2n + 1)) }{ ({3 }^((2n + 2))) \sdot( {3}^((2n - 4)) )}`

let's break it up :

• 
  `\sf  3 {}^(2n) = (3 {}^(2n -4) ) \sdot(3 {}^(4) )`
• 
  `\sf  3 {}^((2n + 1)) = (3 {}^(2n -4) ) \sdot(3 {}^(5) )`
• 
  `\sf  3 {}^((2n + 2)) = (3 {}^(2n -4) ) \sdot(3 {}^(6) )`

now let's take
`{ \sf {3}^((2n-4))}` common here ~

`\qquad \sf  \dashrightarrow \: \frac{(3{)}^((2n - 4)) (3 {}^(4) \sdot{3}^(5)) }{ {(3 )}^((2n - 4))(3 {}^(6) \sdot1)}`

`\qquad \sf  \dashrightarrow \: \frac{{} (3 {}^(4) \sdot{3}^(5)) }{ (3 {}^(6) \sdot1)}`

`\qquad \sf  \dashrightarrow \: \frac{{} 3 {}^(9) }{ 3 {}^(6) }`

`\qquad \sf  \dashrightarrow \: 3 {}^(3)`

`\qquad \sf  \dashrightarrow \: 27`

Answer 2

• 27

Explanation:

`\sf \cfrac{243^{(2n)/(5)}\cdot \:3^(2n+1)}{9^(n+1)\cdot \:3^(2\left(n-2\right))}`

`\sf 9^(n+1)\cdot \:3^(2\left(n-2\right))`

Simplify:-

• 
  `\sf 3^(4n-2)`

`\sf \cfrac{3^(2n)\cdot \:3^(2n+1)}{\boxed{\bf 3^(4n-2)}}`

Now, Factor:-

• 
  `\sf 243^{(2n)/(5)}`

• 
  `= \boxed{\bf 3^(2n)}`

`\sf \cfrac{3^(2n)\cdot \:3^(2n+1)}{3^(4n-2)}`

Now, let's simplify:-

• 
  `\sf \cfrac{3^(2n)}{3^(4n-2)}`

• 
  `=\boxed{\bf 3^(-2n+2)}`

• 
  `\sf 3^(-2n+2)\cdot \:3^(2n+1)`

Simplify:-

• 
  `\sf 3^(-2n+2)\cdot \:3^(2n+1)`

Apply the exponent rule:-

• 
  `\sf 3^(-2n+2+2n+1)`

• 
  `\sf 3^3`

• 
  `\sf 27`

Therefore, your answer is 27.

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Hope this helps!

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