please answer the questiongive the answer fast as possible don't give the bakwas please
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wrong question ...2 is unnecessary in this question ..
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Rahika1:
it is correct answer it is come in exam
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[9^n x 3^2 x (3^(-n/2))^2 - (27)^n ] / 3^3m x 2^3 = 1/27
9^n = 3^2n... (3^(-n/2))^-2 = 3^n... 27^n = 3^3n
Hence, [3^2n x 3^2 x 3^n - 3^3n ]/ 3^3m x 2^3 = 1/27
[3^(3n) x 3^2 - 3^(3n) ]/ 3^3m x 2^3 = 1/27
3^3n [ 3^2 - 1]/3^3m x 2^3 = 1/(3^3)
[3^3n x 8 ] / 3^3m x 8 = 3^(-3)
3^3n/3^3m = 3^(-3)
3^(3n-3m) = 3^(-3)
Hence, 3n - 3m = -3
n - m = -1
m - n = 1...Hence proved
Hope it helps(though little late)..
9^n = 3^2n... (3^(-n/2))^-2 = 3^n... 27^n = 3^3n
Hence, [3^2n x 3^2 x 3^n - 3^3n ]/ 3^3m x 2^3 = 1/27
[3^(3n) x 3^2 - 3^(3n) ]/ 3^3m x 2^3 = 1/27
3^3n [ 3^2 - 1]/3^3m x 2^3 = 1/(3^3)
[3^3n x 8 ] / 3^3m x 8 = 3^(-3)
3^3n/3^3m = 3^(-3)
3^(3n-3m) = 3^(-3)
Hence, 3n - 3m = -3
n - m = -1
m - n = 1...Hence proved
Hope it helps(though little late)..
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