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Difference between revisions of "2008 AMC 10B Problems/Problem 3"
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<math>\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</math>
 
<math>\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</math>
  
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==Solution==
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==Solution 1==
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<math>\sqrt[3]{x\sqrt{x}}=\sqrt[3]{\sqrt{x^3}}=\sqrt[6]{x^3}=x^{3/6}=x^{1/2}\ \mathrm{(D)}</math>
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<math>\sqrt[3]{x\sqrt{x}}=\sqrt[3]{\sqrt{x^3}}=\sqrt[6]{x^3}=x^{3/6}=x^{1/2}\ \boxed{(D)}</math>
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==Solution 2==
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Let x = 64. Using substitution, <math>\sqrt[3]{64\sqrt{64}} = \sqrt[3]{64 \cdot 8} = \sqrt[3]{512} = 8 = \sqrt{64} = 64^\frac{1}{2} = x^\frac{1}{2}</math>, so the answer is <math> \boxed{\textbf{(D)}\ x^\frac{1}{2}}</math>
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~idk12345678
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2008|ab=B|num-b=2|num-a=4}}
 
{{AMC10 box|year=2008|ab=B|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 11:54, 12 June 2024

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 (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

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