Difference between revisions of "2008 AMC 10B Problems/Problem 3"
Megaboy6679 (talk | contribs) m |
Idk12345678 (talk | contribs) |
||
Line 4: | Line 4: | ||
<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}\ \boxed{(D)}</math> | <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}} | {{MAA Notice}} |
Revision as of 18:54, 24 April 2024
Contents
[hide]Problem
Assume that is a positive real number. Which is equivalent to ?
Solution 1
Solution 2
Let x = 64. Using substitution, , 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.