2008 AMC 10B Problems/Problem 3

Problem

Assume that $x$ is a positive real number. Which is equivalent to $\sqrt[3]{x\sqrt{x}}$?

$\mathrm{(A)}\ x^{1/6}\qquad\mathrm{(B)}\ x^{1/4}\qquad\mathrm{(C)}\ x^{3/8}\qquad\mathrm{(D)}\ x^{1/2}\qquad\mathrm{(E)}\ x$

Solution 1

$\sqrt[3]{x\sqrt{x}}=\sqrt[3]{\sqrt{x^3}}=\sqrt[6]{x^3}=x^{3/6}=x^{1/2}\ \boxed{(D)}$

Solution 2

Let x = 64. Using substitution, $\sqrt[3]{64\sqrt{64}} = \sqrt[3]{64 \cdot 8} = \sqrt[3]{512} = 8 = \sqrt{64} = 64^\frac{1}{2} = x^\frac{1}{2}$, so the answer is $\boxed{\textbf{(D)}\ x^\frac{1}{2}}$

~idk12345678

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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All AMC 10 Problems and Solutions

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