What is 2*\sqrt[n]{2^{n-4}}*\sqrt[n]{4^{n+2}} ?

The solution to 2*\sqrt[n]{2^{n-4}}*\sqrt[n]{4^{n+2}} is 16

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simplify 2\sqrt[n]{2^{n-4}}\sqrt[n]{4^{n+2}}

simplify

16

Solution steps

Apply radical rule:

assuming

a \geq 0, b \geq 0

= 2 (2^{n - 4} \cdot 4^{n + 2})^{\frac{1}{n}}

Simplify

2^{n - 4} \cdot 4^{n + 2} : 8^{n}

= 2 (8^{n})^{\frac{1}{n}}

(8^{n})^{\frac{1}{n}} = 8^{1}

= 2 \cdot 8^{1}

Apply rule

a^{1} = a

8^{1} = 8

= 2 \cdot 8

Multiply the numbers:

2 \cdot 8 = 16

= 16

5.67 \cdot 1 0^{- 8} \cdot 534 4^{4}

e^{- \frac{2 x}{y}}

3.65 \cdot \frac{35.575}{5}

x e^{\frac{x}{2}} \cdot e^{\frac{x}{2}} 1 - x^{2}

(9 x^{3} y^{4} z^{3}) (4 x^{2} y^{3} z)

What is 2*\sqrt[n]{2^{n-4}}*\sqrt[n]{4^{n+2}} ?
The solution to 2*\sqrt[n]{2^{n-4}}*\sqrt[n]{4^{n+2}} is 16