If $$m=\frac{2^{5}\times (3^{5})^{2}}{12\times 81\times 3^{3}},$$​ $$n=\frac{(2^{-4})^{2}\times 2^{-5}}{4^{-3}}$$​then what is the value of $$\sqrt{mn}$$​ ?

Correct option is D

​Given :

​$$m=\frac{2^{5}\times (3^{5})^{2}}{12\times 81\times 3^{3}}, \quad$$​​

​$$n=\frac{(2^{-4})^{2}\times 2^{-5}}{4^{-3}}$$​​
Formula Used :

($$a^{m})^{n}=a^{mn}$$​​

​$$a^{m}\div a^{n}=a^{m-n}$$​​

Solution :

m=$$\frac{2^{5}\times 3^{10}}{2^{2}\times 3^{8}}$$​​
​$$=2^{3}\times 3^{2}$$​​
=$$8\times 9$$​​
=72

n$$=\frac{2^{-8}\times 2^{-5}}{(2^{2})^{-3}}$$​​

​$$=\frac{2^{-13}}{2^{-6}}$$​​
​$$=2^{-7}$$​​
​$$=\frac{1}{128}$$​​

mn=$$72\times \frac{1}{128}$$​​
​$$=\frac{9}{16}$$​​
​$$\sqrt{mn}=\sqrt{\frac{9}{16}}=\frac{3}{4}$$​​