Question

Simplify and write the following in exponential form : $\frac{{\left(-6\right)}^2\times 3^{-4}\times 12}{9\times {\left(3^2\right)}^3}$

Answer 1

Given that , : $\frac{{\left(-6\right)}^2\times 3^{-4}\times 12}{9\times {\left(3^2\right)}^3}$

$=\frac{{\left(-6\right)}^2\times {\left(3\right)}^{-4}\times 2^2\times 3}{3^2\times {\left(3\right)}^{3\times 2}}$

$=\frac{{\left(-6\right)}^2\times {\left(3\right)}^{-4}\times 2^2\times 3}{3^2\times {\left(3\right)}^6}$

$=\frac{{\left(-6\right)}^2\times 2^2\times {\left(3\right)}^{-4+1}}{3^{2+6}}$

$=\frac{{\left(-6\right)}^2\times 2^2\times {\left(3\right)}^{-3}}{3^8}$

$={\left(-6\right)}^2\times 2^2\times {\left(3\right)}^{-3}\times 3^{-8}$

$={\left(-12\right)}^2\times 3^{-11}$

$=\frac{{\left(-12\right)}^2}{3^{11}}$

$=\frac{2^4}{3^9}$

Hence , the required value is $\frac{2^4}{3^9}$.