Given that \(81\times 2^{2n-2} = K, find \sqrt{K}\)

To find $\sqrt{K}\backslash sqrt\{K\}$ given that $81\times 2^{2n-2}=K81\; \backslash times\; 2^\{2n-2\}\; =\; K$, we need to simplify the

expression step-by-step.

First,

rewrite $8181$ in terms of powers of $33$: $81=3^{4}81\; =\; 3^4$

So,

the given equation becomes: $K=3^4\times 2^{2n-2}K\; =\; 3^4\; \backslash times\; 2^\{2n-2\}$

Next,

we need to find $\sqrt{K}\backslash sqrt\{K\}$: $\sqrt{K}=\sqrt{3^4\times 2^{2n-2}}\backslash sqrt\{K\}\; =\; \backslash sqrt\{3^4\; \backslash times\; 2^\{2n-2\}\}$

Using the property of square roots $\sqrt{a\times b}=\sqrt{a}\times \sqrt{b}\backslash sqrt\{a\; \backslash times\; b\}\; =\; \backslash sqrt\{a\}\; \backslash times\; \backslash sqrt\{b\}$: $\sqrt{K}=\sqrt{3^4}\times \sqrt{2^{2n-2}}\backslash sqrt\{K\}\; =\; \backslash sqrt\{3^4\}\; \backslash times\; \backslash sqrt\{2^\{2n-2\}\}$

Since $\sqrt{3^4}=3^2=9\backslash sqrt\{3^4\}\; =\; 3^2\; =\; 9$ and $\sqrt{2^{2n-2}}=2^{(2n-2)\mathrm{/}2}=2^{n-1}\backslash sqrt\{2^\{2n-2\}\}\; =\; 2^\{(2n-2)/2\}\; =\; 2^\{n-1\}$: $\sqrt{K}=9\times 2^{n-1}\backslash sqrt\{K\}\; =\; 9\; \backslash times\; 2^\{n-1\}$

Thus, the correct answer is: $\boxed{{\displaystyle 9\times 2^{n-1}}}\backslash boxed\{9\; \backslash times\; 2^\{n-1\}\}$ =