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

e^{- \frac{x}{a}}

\frac{( 1 + a x ) ^{2}}{( 1 - x ) ^{2}}

e^{- l n ∣ 1 + k (1 - t) ∣}

1 c K^{6} KH Y^{6} m H g Y v m r Q^{4} P aa f K Y Dr g^{1} e jb H^{1} ce

7 e^{- (\frac{1}{2})^{3}}

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