What is \sqrt[n]{(20)/(2^{2n+4)+2^{2n+2}}} ?

The solution to \sqrt[n]{(20)/(2^{2n+4)+2^{2n+2}}} is \sqrt[n]{1/(4^n)}

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simplify \sqrt[n]{(20)/(2^{2n+4)+2^{2n+2}}}

simplify

Solution steps

Factor

2^{2 n + 4} + 2^{2 n + 2} : 2^{2 n + 2} \cdot 5

Factor the number:

20 = 5 \cdot 4

Cancel the common factor:

5

Factor the number:

4 = 2^{2}

Simplify

\frac{2 ^{2}}{2 ^{2 n + 2}} : \frac{1}{2 ^{2 n}}

Simplify

2^{2 n} : 4^{n}

\frac{5 ^{- 2}}{5 ^{- 3}}

(2 q + r)^{2}

100 \div 50

14 \cdot 14

(a^{3})^{3}

What is \sqrt[n]{(20)/(2^{2n+4)+2^{2n+2}}} ?
The solution to \sqrt[n]{(20)/(2^{2n+4)+2^{2n+2}}} is \sqrt[n]{1/(4^n)}