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> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | <math>\sqrt[3]{x\sqrt{x}}=\sqrt[3]{\sqrt{x^3}}=\sqrt[6]{x^3}=x^{3/6}=x^{1/2}\ \ | + | <math>\sqrt[3]{x\sqrt{x}}=\sqrt[3]{\sqrt{x^3}}=\sqrt[6]{x^3}=x^{3/6}=x^{1/2}\ \boxed{(D)}</math> |
| + | |||
| + | ==Solution 2== | ||
| + | Let x = 64. Using substitution, <math>\sqrt[3]{64\sqrt{64}} = \sqrt[3]{64 \cdot 8} = \sqrt[3]{512} = 8 = \sqrt{64} = 8^\frac{1}{2} = x^\frac{1}{2}</math>, so the answer is <math> \boxed{\textbf{(D)}\ x^\frac{1}{2}}</math> | ||
| + | |||
| + | ~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}} | ||
Latest revision as of 19:54, 24 April 2024
Contents
Problem
Assume that 
is a positive real number. Which is equivalent to ![$\sqrt[3]{x\sqrt{x}}$](http://latex.artofproblemsolving.com/b/b/6/bb6b6aec07627f5fe5db310965f0a81771c5a48e.png)
?
Solution 1
![$\sqrt[3]{x\sqrt{x}}=\sqrt[3]{\sqrt{x^3}}=\sqrt[6]{x^3}=x^{3/6}=x^{1/2}\ \boxed{(D)}$](http://latex.artofproblemsolving.com/0/b/9/0b9835f84d0ba8389cc915b1632ecf1d80034b10.png)
Solution 2
Let x = 64. Using substitution, ![$\sqrt[3]{64\sqrt{64}} = \sqrt[3]{64 \cdot 8} = \sqrt[3]{512} = 8 = \sqrt{64} = 8^\frac{1}{2} = x^\frac{1}{2}$](http://latex.artofproblemsolving.com/4/e/e/4eeab609ed276ab7a813790482ba4a3963f501dd.png)
, so the answer is 
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
