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 ===
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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>
  
=== Solution 2 ===
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== 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??)===
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==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 \boxed{(A) \sqrt{x}}$.
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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>.
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 +
~Darth_Cadet
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 +
==Video Solution==
 +
 
 +
https://www.youtube.com/watch?v=8DMxo2pi1h8  ~David
  
 
== See Also ==
 
== See Also ==

Latest revision as of 17:47, 28 July 2023

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 $x^\frac{1}{2} = \boxed{(A) \sqrt{x}}$.

Solution 3 (Lame??)

WLOG, plug in some random square number like $9$ or $4$, and the output you get after simplifying the expression will always be the square root of the number, so the answer is just $\boxed{(A) \sqrt{x}}$.

~Darth_Cadet

Video Solution

https://www.youtube.com/watch?v=8DMxo2pi1h8 ~David

See Also

2003 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AMC 10 Problems and Solutions

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