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

Latest revision as of 13:28, 5 July 2013

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 $\boxed{D}$

1998 AHSME (Problems • Answer Key • Resources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 5: / Line 5: @@
 ==Solution==
-{{solution}}
+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>
+<math>\sqrt[3]{N\sqrt[3]{N\sqrt[3]{N}}}</math>
+<math>\sqrt[3]{N\sqrt[3]{N\cdot N^{\frac{1}{3}}}}</math>
+<math>\sqrt[3]{N\sqrt[3]{N^{\frac{4}{3}}}}</math>
+<math>\sqrt[3]{N\cdot{N^{\frac{4}{9}}}}</math>
+<math>\sqrt[3]{{N^{\frac{13}{9}}}}</math>
+<math>N^{\frac{13}{27}}</math>, thus the answer is <math>\boxed{D}</math>
 ==See Also==
 {{AHSME box|year=1998|num-b=6|num-a=8}}
+{{MAA Notice}}

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

Latest revision as of 13:28, 5 July 2013

Problem

Solution

See Also