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

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==Solution==
 
==Solution==
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The key identities are <math>\sqrt[3]{x^n} = x^{\frac{n}{3}}</math> and <math>x \cdot x^{\frac{a}{b}} = x^{1 + \frac{a}{b}}</math>
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<math>\sqrt[3]{N\sqrt[3]{N\sqrt[3]{N}}}</math>
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<math>\sqrt[3]{N\sqrt[3]{N\cdot N^{\frac{1}{3}}}}</math>
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<math>\sqrt[3]{N\sqrt[3]{N^{\frac{4}{3}}}}</math>
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<math>\sqrt[3]{N\cdot{N^{\frac{4}{9}}}}</math>
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<math>\sqrt[3]{{N^{\frac{13}{9}}}}</math>
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<math>N^{\frac{13}{27}}</math>, thus the answer is <math>\boxed{D}</math>
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==See Also==
 
==See Also==
 
{{AHSME box|year=1998|num-b=6|num-a=8}}
 
{{AHSME box|year=1998|num-b=6|num-a=8}}

Revision as of 21:46, 7 August 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

The key identities are $\sqrt[3]{x^n} = x^{\frac{n}{3}}$ and $x \cdot x^{\frac{a}{b}} = x^{1 + \frac{a}{b}}$

$\sqrt[3]{N\sqrt[3]{N\sqrt[3]{N}}}$

$\sqrt[3]{N\sqrt[3]{N\cdot N^{\frac{1}{3}}}}$

$\sqrt[3]{N\sqrt[3]{N^{\frac{4}{3}}}}$

$\sqrt[3]{N\cdot{N^{\frac{4}{9}}}}$

$\sqrt[3]{{N^{\frac{13}{9}}}}$

$N^{\frac{13}{27}}$, thus the answer is $\boxed{D}$

See Also

1998 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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 26 27 28 29 30
All AHSME Problems and Solutions