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 $x^{\frac{7}{8}}$ .

1991 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1991 AHSME Problems/Problem 6

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