Question

Which is equivalent to RootIndex 3 StartRoot 8 EndRoot Superscript one-fourth x?

8 Superscript three-fourths x
RootIndex 7 StartRoot 8 EndRoot Superscript x
RootIndex 12 StartRoot 8 EndRoot Superscript x
8 Superscript StartFraction 3 Over 4 x EndFraction

Answer 1

C 8 x/3

Step-by-step explanation:

Answer 2

Answer: Choice C

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)`

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Step-by-step explanation:

The general rule is

`\sqrt[n]{x} = x^(1/n)`

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)`

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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,

`\left(8^(1/3)\right)^((1/4)x) = 8^((1/3)*(1/4)x)`

`\left(8^(1/3)\right)^((1/4)x) = 8^((1/12)x)`

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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)}`

`8^((1/12)x) = \left(\sqrt[12]{8}\right)^(x)`