1998 AHSME Problems/Problem 7

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1998 AHSME Problems/Problem 7

Problem

Solution

See Also