Question

If $\frac{9^n\times 3^5\times {\left(27\right)}^3}{3\times {\left(81\right)}^4}=27$, then the value of $n$ is

• A. $0$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (C)

$3$

$\frac{9^n\times 3^5\times (27{)}^3}{3\times (81{)}^4}=27$

$\Rightarrow$ $\frac{3^{2n}\times 3^5\times (3^3{)}^3}{3^1\times (3^4{)}^4}=3^3$

$\Rightarrow$ $\frac{3^{2n}\times 3^5\times 3^9}{3^1\times 3^{16}}=3^3$

$\Rightarrow$ $\frac{3^{2n}\times 3^{5+9}}{3^{1+16}}=3^3$ ....$[a^m\times a^n=a^{m+n}]$

$\Rightarrow$ $\frac{3^{2n}\times 3^{14}}{3^{17}}=3^3$

$\Rightarrow$ $3^{2n}=\frac{3^3\times 3^{17}}{3^{14}}$

$\Rightarrow$ $3^{2n}=\frac{3^{3+17}}{3^{14}}$

$\Rightarrow$ $3^{2n}=\frac{3^{20}}{3^{14}}$

$\Rightarrow$ $3^{2n}=3^{20-14}$

$\Rightarrow$ $3^{2n}=3^6$

$\therefore$ $2n=6$

$\therefore$ $n=3$