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

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

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

simplify

16

Solution steps

Apply radical rule:

assuming

a \geq 0, b \geq 0

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

Simplify

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

(\frac{- 5 + 37}{2})^{3} \cdot (\frac{- 5 - 37}{2})^{3}

(4 x) (4 x^{2})

- 6 e^{- x} cos (x)

e^{- 1.14}

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