Question

Simplify and write the following in exponential form : $\left[{\left(2\right)}^{-3}÷{\left(2\right)}^{-8}\right]\times \left[{\left(3^2\right)}^{-2}÷{\left(3\right)}^{-4}\right]$ .

Answer 1

Given that , $\left[{\left(2\right)}^{-3}÷{\left(2\right)}^{-8}\right]\times \left[{\left(3^2\right)}^{-2}÷{\left(3\right)}^{-4}\right]$ .

$={\left(2\right)}^{-3}\times 2^8\times 3^{2\times -2}\times 3^4$

$=2^{-3}\times 2^8\times 3^{-4}\times 3^4$

$=2^{-3+8}\times 3^{-4+4}$

$=2^5\times 3^0$

$=2^5\times 1$

$=2^5$

$=32$

Hence , the required value is $2^{6}or32$ .