Question

Prove that m-n =1 when 9 power n*3 square * (3power -n/2)power -2 -(27)powern whole divided by 3power 3m * 2power 3 =1/27

Answer 1

Answer:

•]••´º´•» Hey «•´º´••[•

Step-by-step explanation:

Given that, ( 9^n x 3^2 x (3^-n/2)^-2 - 27^n) /( 3^3m x 2^3) = 1/27

⇒ (3^2n x 3^2 x { 3^-(-2n/2) } - 3^3n )/ 3^3m x 2^3 = 1/(3^3)

⇒ (3^2n x 3^2 x {3^n} - 3^3n )/ 3^3m x 2^3 = 3^(-3)

⇒ ( 3^3n x 3^2 - 3^3n )/ 3^3m x 2^3 = 3^(-3)

⇒ 3^3n[ 3^2 - 1] / 3^3m x 2^3 = 3^(-3)

⇒ 3^3n[ 9 - 1 ] /3^3m x 8 =  3^(-3)

⇒ 3^3n x 8 /3^3m x 8 = 3^(-3)

 ⇒ 3^(3n-3m) = 3^(-3)

comparing powers on both side, we get

3n - 3m = -3

⇒ 3(n-m) = -3

⇒ (n-m) = -1

or m-n = 1

Hence proved

↫↫↫↫↫ Thx.. ↬↬↬↬↬

Answer 2

Given that, ( 9^n x 3^2 x (3^-n/2)^-2 - 27^n) /( 3^3m x 2^3) = 1/27

⇒ (3^2n x 3^2 x { 3^-(-2n/2) } - 3^3n )/ 3^3m x 2^3 = 1/(3^3)

⇒ (3^2n x 3^2 x {3^n} - 3^3n )/ 3^3m x 2^3 = 3^(-3)

⇒ ( 3^3n x 3^2 - 3^3n )/ 3^3m x 2^3 = 3^(-3)

⇒ 3^3n[ 3^2 - 1] / 3^3m x 2^3 = 3^(-3)

⇒ 3^3n[ 9 - 1 ] /3^3m x 8 =  3^(-3)

⇒ 3^3n x 8 /3^3m x 8 = 3^(-3)

 ⇒ 3^(3n-3m) = 3^(-3)

comparing powers on both side, we get

3n - 3m = -3

⇒ 3(n-m) = -3

⇒ (n-m) = -1

or m-n = 1

Hence proved

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