Question

If $\frac{9^{n+1}\times (3^{\frac{n}{2}}{)}^{-2}\times {27}^n}{(3^m{)}^4}=\frac{1}{729}$, then $m-n$ equals to.

• A. $2$
• B. $1$
• C. $3$
• D. $4$

Answer 1

Answer: (A)

$2$

Given,

$\frac{9^{n+1}\times (3^{\frac{n}{2}}{)}^{-2}\times {27}^n}{(3^m{)}^4}=\frac{1}{729}$

or, $\frac{3^{2n+2}\times \left(3^{-n}\right)\times 3^{3n}}{\left(3^{4m}\right)}=\frac{1}{9^3}$

or, $\frac{3^{4n+2}}{\left(3^{4m}\right)}=\frac{1}{3^6}$

or, $3^{4(n-m)}\times 3^2=\frac{1}{3^6}$

or, $3^{4(n-m)}=3^{-8}$

or, $4(n-m)=-8$

or, $m-n=2$.