Math, asked by sethjoanna012, 1 year ago

If( 9^n x 3^2 x (3^-n/2)^-2 - 27^n) /( 3^3m x 2^3) = 1/27, then prove that m-n=1.

Answers

Answered by Aaeesha
276
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

sethjoanna012: thank you so much friend!!
Answered by navadeep7
45
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

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