Evalute $\sqrt[n]{\frac{20}{2^{2n+4}+2^{2n+2}}}$

Question

Evaluate the given expression $$\sqrt[n]{\dfrac{20}{2^{2n+4}+2^{2n+2}}}$$ The given answer is $\dfrac{1}{4}$. My attempt: $$\sqrt[n]{\dfrac{20}{2^{2n+4}+2^{2n+2}}}=\sqrt[n]{\dfrac{20}{2^{2n}\cdot2^4+2^{2n}\cdot2}}\\=\sqrt[n]{\dfrac{20}{2^{2n}\cdot18}}=\sqrt[n]{\dfrac{10}{9\cdot2^{2n}}}$$ This is as far I as I am able to reach. Thank you!

PS I don't see how one can get $\dfrac14$. For that we have to get something like $\sqrt[n]{A^n}$.

Answer 1

Since $2^{2n+4}+2^{2n+2}=2^{2n}(2^4+2^2)=20\times4^n$,$$\sqrt[n]{\frac{20}{2^{2n+4}+2^{2n+2}}}=\sqrt[n]{\frac{20}{20\times4^n}}=\frac14.$$

Answer 2

You made a small mistake :\begin{aligned}\sqrt[n]{\dfrac{20}{2^{2n+4}+2^{2n+2}}}&=\sqrt[n]{\dfrac{20}{2^{2n}\cdot2^4+2^{2n}\cdot2^{\color{red}{2}}}}\\&=\sqrt[n]{\dfrac{20}{2^{2n}\cdot\color{red}{20}}}=\sqrt[n]{\dfrac{1}{2^{2n}}}=\frac{1}{4}\end{aligned}

Answer 3

Using the identity $\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$, we find that $$ \begin{align*} \sqrt[n]{\frac{20}{2^{2n+4}+2^{2n+2}}} &= \sqrt[n]{\frac1{2^n}}\cdot\sqrt[n]{\frac{20}{2^4+2^2}}\\ &= \sqrt[n]{\frac1{4^n}}\cdot\sqrt[n]{\frac{20}{16+4}}\\ &=\sqrt[n]{\left(\frac14\right)^n}\cdot\sqrt[n]{\frac{20}{20}}\\ &=\frac14. \end{align*} $$