Question

The value of $\frac{{\left(27\right)}^{\frac{n}{3}}\times {\left(8\right)}^{-\frac{n}{6}}}{{\left(162\right)}^{-\frac{n}{2}}}$ is equal to

• A. $2^n$
• B. $3^n$
• C. ${27}^n$
• D. ${35}^n$

Answer 1

The correct option is C ${27}^n$

$Now\frac{{\left(27\right)}^{\frac{n}{3}}\times {\left(8\right)}^{-\frac{n}{6}}}{{\left(162\right)}^{-\frac{n}{2}}}\to \frac{{\left({\left(3\right)}^3\right)}^{\frac{n}{3}}\times {\left({\left(2\right)}^3\right)}^{-\frac{n}{6}}}{{(81\times 2)}^{-\frac{n}{2}}}\to \frac{{\left(3\right)}^{3\times \frac{n}{3}}\times {\left(2\right)}^{-3\times \frac{n}{6}}}{{(3^4\times 2)}^{-\frac{n}{2}}}\to \frac{{\left(3\right)}^n\times {\left(2\right)}^{-\frac{n}{2}}}{(3^{4\times -\frac{n}{2}}\times 2^{-\frac{n}{2}})}\to \frac{{\left(3\right)}^n}{\left(3^{-2n}\right)}=\frac{3^n}{3^{-2n}}=3^{3n}={27}^n$