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 ==
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== 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>
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== Solution 2 ==
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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>.
  
 
== See Also ==
 
== See Also ==

Revision as of 23:20, 9 June 2015

Problem

Simplify $\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}$.

$\mathrm{(A) \ } \sqrt{x}\qquad \mathrm{(B) \ } \sqrt[3]{x^{2}}\qquad \mathrm{(C) \ } \sqrt[27]{x^{2}}\qquad \mathrm{(D) \ } \sqrt[54]{x}\qquad \mathrm{(E) \ } \sqrt[81]{x^{80}}$

Solution 1

$\sqrt[3]{x\sqrt{x}}=\sqrt[3]{(\sqrt{x})^{2}\cdot\sqrt{x}}=\sqrt[3]{(\sqrt{x})^{3}}=\sqrt{x}$.

Therefore:

$\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}=\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}=\sqrt[3]{x\sqrt{x}}= \sqrt{x} \Rightarrow \boxed{\mathrm{(A)}\ \sqrt{x}}$

Solution 2

We know that $\sqrt[3]{x\sqrt{x}} = \sqrt[3]{x\cdot(x^\frac{1}{2})} = \sqrt[3]{x^\frac{3}{2}} = x^\frac{1}{2}$ We plug this into $\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}$ and get $\sqrt[3]{x\sqrt[3]{x\cdot(x^\frac{1}{2})}}$. Again, we substitute and get $\sqrt[3]{x\cdot(x^\frac{1}{2})}$. We substitute one more time and get $x^\frac{1}{2} = \boxed{(A) \sqrt{x}}$.

See Also

2003 AMC 10A (ProblemsAnswer KeyResources)
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

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