Difference between revisions of "2003 AMC 10A Problems/Problem 9"
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<math> \mathrm{(A) \ } \sqrt{x}\qquad \mathrm{(B) \ } \sqrt[3]{x^{2}}\qquad \mathrm{(C) \ } \sqrt[27]{x^{2}}\qquad \mathrm{(D) \ } \sqrt[54]{x}\qquad \mathrm{(E) \ } \sqrt[81]{x^{80}} </math> | <math> \mathrm{(A) \ } \sqrt{x}\qquad \mathrm{(B) \ } \sqrt[3]{x^{2}}\qquad \mathrm{(C) \ } \sqrt[27]{x^{2}}\qquad \mathrm{(D) \ } \sqrt[54]{x}\qquad \mathrm{(E) \ } \sqrt[81]{x^{80}} </math> | ||
| − | + | == Solution 1 == | |
<math>\sqrt[3]{x\sqrt{x}}=\sqrt[3]{(\sqrt{x})^{2}\cdot\sqrt{x}}=\sqrt[3]{(\sqrt{x})^{3}}=\sqrt{x}</math>. | <math>\sqrt[3]{x\sqrt{x}}=\sqrt[3]{(\sqrt{x})^{2}\cdot\sqrt{x}}=\sqrt[3]{(\sqrt{x})^{3}}=\sqrt{x}</math>. | ||
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<math>\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}=\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}=\sqrt[3]{x\sqrt{x}}= \sqrt{x} \Rightarrow \boxed{\mathrm{(A)}\ \sqrt{x}}</math> | <math>\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}=\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}=\sqrt[3]{x\sqrt{x}}= \sqrt{x} \Rightarrow \boxed{\mathrm{(A)}\ \sqrt{x}}</math> | ||
| − | + | == Solution 2 == | |
We know that <math>\sqrt[3]{x\sqrt{x}} = \sqrt[3]{x\cdot(x^\frac{1}{2})} = \sqrt[3]{x^\frac{3}{2}} = x^\frac{1}{2}</math> We plug this into <math>\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}</math> and get <math>\sqrt[3]{x\sqrt[3]{x\cdot(x^\frac{1}{2})}}</math>. Again, we substitute and get <math>\sqrt[3]{x\cdot(x^\frac{1}{2})}</math>. We substitute one more time and get <math>x^\frac{1}{2} = \boxed{(A) \sqrt{x}}</math>. | We know that <math>\sqrt[3]{x\sqrt{x}} = \sqrt[3]{x\cdot(x^\frac{1}{2})} = \sqrt[3]{x^\frac{3}{2}} = x^\frac{1}{2}</math> We plug this into <math>\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}</math> and get <math>\sqrt[3]{x\sqrt[3]{x\cdot(x^\frac{1}{2})}}</math>. Again, we substitute and get <math>\sqrt[3]{x\cdot(x^\frac{1}{2})}</math>. We substitute one more time and get <math>x^\frac{1}{2} = \boxed{(A) \sqrt{x}}</math>. | ||
| − | + | ==Solution 3 (Lame??)== | |
WLOG, plug in some random square number like <math> 9 </math> or <math> 4 </math>, and the output you get after simplifying the expression will always be the square root of the number, so the answer is just <math> \boxed{(A) \sqrt{x}}</math>. | WLOG, plug in some random square number like <math> 9 </math> or <math> 4 </math>, and the output you get after simplifying the expression will always be the square root of the number, so the answer is just <math> \boxed{(A) \sqrt{x}}</math>. | ||
Latest revision as of 17:47, 28 July 2023
Problem
Simplify
![$\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}$](http://latex.artofproblemsolving.com/c/f/2/cf2756b7e9bb76b2adf76f1709cc765535b27602.png)
.
![$\mathrm{(A) \ } \sqrt{x}\qquad \mathrm{(B) \ } \sqrt[3]{x^{2}}\qquad \mathrm{(C) \ } \sqrt[27]{x^{2}}\qquad \mathrm{(D) \ } \sqrt[54]{x}\qquad \mathrm{(E) \ } \sqrt[81]{x^{80}}$](http://latex.artofproblemsolving.com/2/e/1/2e12822d4d418ef37b7ca03cff39d5b6e8123508.png)
Solution 1
![$\sqrt[3]{x\sqrt{x}}=\sqrt[3]{(\sqrt{x})^{2}\cdot\sqrt{x}}=\sqrt[3]{(\sqrt{x})^{3}}=\sqrt{x}$](http://latex.artofproblemsolving.com/5/b/5/5b532d915bdc855d96cfa266d7d3bb9a168fd803.png)
.
Therefore:
![$\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}=\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}=\sqrt[3]{x\sqrt{x}}= \sqrt{x} \Rightarrow \boxed{\mathrm{(A)}\ \sqrt{x}}$](http://latex.artofproblemsolving.com/7/a/6/7a6a1831eb969273f9c00ce0ca3ba2e52d2b5c72.png)
Solution 2
We know that ![$\sqrt[3]{x\sqrt{x}} = \sqrt[3]{x\cdot(x^\frac{1}{2})} = \sqrt[3]{x^\frac{3}{2}} = x^\frac{1}{2}$](http://latex.artofproblemsolving.com/2/7/9/279577b98d2f0150723f2f0515956c7644b64f25.png)
We plug this into and get ![$\sqrt[3]{x\sqrt[3]{x\cdot(x^\frac{1}{2})}}$](http://latex.artofproblemsolving.com/0/6/a/06a99fffd0d716670d45d3670d720ce5a2759b5e.png)
. Again, we substitute and get ![$\sqrt[3]{x\cdot(x^\frac{1}{2})}$](http://latex.artofproblemsolving.com/1/b/1/1b1adbd26fb38c3122623b411b4ac95ec22a10d0.png)
. We substitute one more time and get 
.
Solution 3 (Lame??)
WLOG, plug in some random square number like 
or 
, and the output you get after simplifying the expression will always be the square root of the number, so the answer is just 
.
~Darth_Cadet
Video Solution
https://www.youtube.com/watch?v=8DMxo2pi1h8 ~David
See Also
| 2003 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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. 
