Question

(9^n × 3^2 ×3^n - 27^n)/3^3m × 2^3 = 1/27 Prove that m - n = 1​

Answer 1

heymate

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Answer 2

Hii Nikkuuuuuuuu.........!!!!!!!

STEP 1 : Make the complex portion of the problem simple

(9^n × 3² × 3^n - 27^n)/(3^3m × 2³)

= (3^2n × 3^2 × 3^n - 3^3n)/(3^3m × 2³)

= 3^{2n+n}×3²-3^3n)/(3^3m × 2³)

= (3^3n × 9 - 3^3n)/(3^3m × 2³)

= {3^3n×(9-1)}/(3^3m × 2³)

= (3^3n × 8)/(3^3m × 8)

= 3^3n/3^3m

= 3^{3n-3m}

STEP 2 : Now, it's in simple form. So, now we can proceed the solution easily.

Now, 3^{3n-3m} = 1/27

=> 3^{3n-3m} × 27 = 1

=> 3^{3n-3m} × 3³ = 1 [Because 27 = 3³]

=> 3^{3n-3m+3} = 3^0 [Because, 3^0 = 1]

Equating powers of same bases, we get :

3n - 3m + 3 = 0

=> 3(n-m+1) = 0

=> n-m+1 = 0/3 = 0

=> n-m = -1

=> -m = -1-n

=> m = -(-1-n) = 1+n

=> m-n = 1 [Proved]