Question

If $\frac{9^n\times 3^5\times (27{)}^3}{3\times (81{)}^4}=27$ then, $n$ is equal to:

• A. $0$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (C)

$3$

$\frac{9^n\times 3^5\times {\left(27\right)}^3}{3\times {\left(81\right)}^4}=27$
$\Rightarrow \frac{3^{2n}\times 3^5\times {\left(3^3\right)}^3}{3\times {\left(3^4\right)}^4}=3^3\Rightarrow \frac{3^{2n}\times 3^5\times 3^9}{3\times 3^{16}}=3^3$$\Rightarrow 3^{2n+14-17}=3^3\Rightarrow 3^{2n-3}=3^3$

$\Rightarrow 2n-3=3\Rightarrow 2n=6\Rightarrow n=3$