Question

If 9^n3^2.3^n - (27)^n/ 3^3m2^3
=1/27,
prove that m = 1 + n.​

Answer 1

$\sf\ \red{Given:-}$

• $\sf\ \dfrac{9^n3^2.3^n – (27)^n}{3^3m2^3}= \dfrac{1}{27}$

$\sf\ \red{To\:Prove:-}$

• $\sf\ m = 1–n$

$\sf\ \red{According\:to\:the\:question:-}$

$\dashrightarrow\: \sf\ \dfrac{ (3^2n× 3^2×[3^–(–2n/2)]–3^3n)}{(3^3m×2^3)} =1(3^3)$

$\dashrightarrow\: \sf\ \dfrac{ (3^2n × 3^2 × [3^n] – 3^3n}{ 3^3m × 2^3} = 3^(–3)$

$\dashrightarrow\: \sf\ \dfrac{ (3^3n × 3^2– 3^3n)}{ 3^3m × 2^3} = 3^(–3)$

$\dashrightarrow\: \sf\ \dfrac{3^3n [3^2 – 1] }{ 3^3m × 2^3} = 3^(–3)$

$\dashrightarrow\: \sf\ \dfrac{3^3n [3^2 – 1] }{ 3^3m × 2^3} = 3^(–3)$

$\dashrightarrow\: \sf\ \dfrac{3^3n [9 – 1] }{ 3^3m × 8} = 3^(–3)$

$\dashrightarrow\: \sf\ \dfrac{ 3^3n × 8 }{ 3^3m × 8} = 3^(–3)$

$\dashrightarrow\: \sf\ 3^(3n–3m) = 3^(–3)$

$\sf\ Comparing\:powers\: on\:both\:side,we \: get$

$\dashrightarrow\: \sf\ 3n – 3m = –3$

$\dashrightarrow\: \sf\ 3(n–m) = –3$

$\dashrightarrow\: \sf\ (n–m) = –1$

Or

$\dashrightarrow\: \sf\ m = 1 + n$

Hence proved.