1991 AHSME Problems/Problem 6

Problem

If $x\geq 0$, then $\sqrt{x\sqrt{x\sqrt{x}}}=$

(A) $x\sqrt{x}$ (B) $x \sqrt[4]{x}$ (C) $\sqrt[8]{x}$ (D) $\sqrt[8]{x^3}$ (E) $\sqrt[8]{x^7}$

Solution

$\fbox{E}$ We have $\sqrt{x} = x^{\frac{1}{2}}, x\sqrt{x} = x^{\frac{3}{2}}$, then square-rooting that gives $x^{\frac{3}{4}}$, then multiplying by $x$ gives $x^{\frac{7}{4}}$, then square-rooting this gives $x^{\frac{7}{8}}$.

See also

1991 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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