Difference between revisions of "1998 AHSME Problems/Problem 7"

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== Problem 7 ==
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== Problem ==
 
If <math>N > 1</math>, then <math>\sqrt[3]{N\sqrt[3]{N\sqrt[3]{N}}} =</math>
 
If <math>N > 1</math>, then <math>\sqrt[3]{N\sqrt[3]{N\sqrt[3]{N}}} =</math>
  
 
<math> \mathrm{(A) \ } N^{\frac 1{27}} \qquad \mathrm{(B) \ } N^{\frac 1{9}} \qquad \mathrm{(C) \ } N^{\frac 1{3}} \qquad \mathrm{(D) \ } N^{\frac {13}{27}} \qquad \mathrm{(E) \ } N</math>
 
<math> \mathrm{(A) \ } N^{\frac 1{27}} \qquad \mathrm{(B) \ } N^{\frac 1{9}} \qquad \mathrm{(C) \ } N^{\frac 1{3}} \qquad \mathrm{(D) \ } N^{\frac {13}{27}} \qquad \mathrm{(E) \ } N</math>
  
[[1998 AHSME Problems/Problem 7|Solution]]
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==Solution==
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{{solution}}
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==See Also==
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{{AHSME box|year=1998|num-b=6|num-a=8}}

Revision as of 15:36, 6 June 2011

Problem

If $N > 1$, then $\sqrt[3]{N\sqrt[3]{N\sqrt[3]{N}}} =$

$\mathrm{(A) \ } N^{\frac 1{27}} \qquad \mathrm{(B) \ } N^{\frac 1{9}} \qquad \mathrm{(C) \ } N^{\frac 1{3}} \qquad \mathrm{(D) \ } N^{\frac {13}{27}} \qquad \mathrm{(E) \ } N$

Solution

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See Also

1998 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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All AHSME Problems and Solutions