Question

If $\frac{9^n\left(3^2\right){\left(3^{-n/2}\right)}^{-2}-{27}^n}{3^{3m}\left(2^3\right)}=\frac{1}{27}$, then

• A. m – n = 2
• B. m – n = 1
• C. m – n = – 2
• D. m – n = – 1
• E. None of these

Answer 1

The correct option is B

m – n = 1

$\begin{array}{c}\frac{9^n\left(3^2\right){\left(3^{-n/2}\right)}^{-2}-{27}^n}{3^{3m}\left(2^3\right)}=\frac{1}{27}\\ \Rightarrow \frac{3^{2n}.3^2.3^{\left(-\frac{n}{2}\right)\times \left(-2\right)}-{\left(3^3\right)}^n}{3^{3m}{.2}^3}=\frac{1}{3^3}\\ \Rightarrow \frac{3^{3n}\left(9-1\right)}{3^{3m.}.8}=\frac{1}{3^3}\\ \Rightarrow 3^{3n-3m}=3^{-3}\Rightarrow 3n-3m=-3\\ m-n=1.\end{array}$