Q:

# Which is equivalent to RootIndex 3 StartRoot 8 EndRoot Superscript one-fourth x?8 Superscript three-fourths xRootIndex 7 StartRoot 8 EndRoot Superscript xRootIndex 12 StartRoot 8 EndRoot Superscript x8 Superscript StartFraction 3 Over 4 x EndFraction

Accepted Solution

A:
Answer: Choice CRootIndex 12 StartRoot 8 EndRoot Superscript x12th root of 8^x = (12th root of 8)^x$$\sqrt[12]{8^{x}} = \left(\sqrt[12]{8}\right)^{x}$$=========================================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)$$\sqrt[3]{8} = 8^{1/3}$$-----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)xIn other words,$$\left(8^{1/3}\right)^{(1/4)x} = 8^{(1/3)*(1/4)x}$$$$\left(8^{1/3}\right)^{(1/4)x} = 8^{(1/12)x}$$-----From here, we rewrite the fractional exponent 1/12 as a 12th root. which leads us to this$$8^{(1/12)x} = \sqrt[12]{8^{x}}$$$$8^{(1/12)x} = \left(\sqrt[12]{8}\right)^{x}$$