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

~idk12345678

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2008 AMC 10B Problems/Problem 3

Contents

Problem

Solution 1

Solution 2

See also