<h3>Answer: Choice C</h3>
RootIndex 12 StartRoot 8 EndRoot Superscript x
12th root of 8^x = (12th root of 8)^x
![\sqrt[12]{8^{x}} = \left(\sqrt[12]{8}\right)^{x}](https://tex.z-dn.net/?f=%5Csqrt%5B12%5D%7B8%5E%7Bx%7D%7D%20%3D%20%5Cleft%28%5Csqrt%5B12%5D%7B8%7D%5Cright%29%5E%7Bx%7D)
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Explanation:
The general rule is
![\sqrt[n]{x} = x^{1/n}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%7D%20%3D%20x%5E%7B1%2Fn%7D)
so any nth root is the same as having a fractional exponent 1/n.
Using that rule we can say the cube root of 8 is equivalent to 8^(1/3)
![\sqrt[3]{8} = 8^{1/3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B8%7D%20%3D%208%5E%7B1%2F3%7D)
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Raising this to the power of (1/4)x will have us multiply the exponents of 1/3 and (1/4)x like so
(1/3)*(1/4)x = (1/12)x
In other words,


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From here, we rewrite the fractional exponent 1/12 as a 12th root. which leads us to this
![8^{(1/12)x} = \sqrt[12]{8^{x}}](https://tex.z-dn.net/?f=8%5E%7B%281%2F12%29x%7D%20%3D%20%5Csqrt%5B12%5D%7B8%5E%7Bx%7D%7D%20)
![8^{(1/12)x} = \left(\sqrt[12]{8}\right)^{x}](https://tex.z-dn.net/?f=8%5E%7B%281%2F12%29x%7D%20%3D%20%5Cleft%28%5Csqrt%5B12%5D%7B8%7D%5Cright%29%5E%7Bx%7D%20)
It might be D in my opinion? But I probably did my work wrong.. sorry!
Sounds like a good deal to go swimming!
Answer:
answer : x = 3±√ 119 / 16
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