1998 AHSME Problem 7

Algebra Level 2

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

N N 1 / 27 N^{1/27} N 1 / 3 N^{1/3} N 13 / 27 N^{13/27}

Daniel Hirschberg by Daniel Hirschberg , 39, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Note that :- N 3 = N 1 3 \sqrt[3]{N} =N^{\frac{1}{3}} Therefore the expression is equal to ( N ( N ( N ) 1 3 ) 1 3 ) 1 3 (N(N(N)^{\frac{1}{3}})^{\frac{1}{3}})^{\frac{1}{3}}

= ( N ( N 4 3 ) 1 3 ) 1 3 = (N(N^{\frac{4}{3}})^{\frac{1}{3}})^{\frac{1}{3}}

= ( N ( N 4 9 ) ) 1 3 = (N(N^{\frac{4}{9}}))^{\frac{1}{3}}

= ( N 13 9 ) 1 3 = (N^{\frac{13}{9}})^{\frac{1}{3}}

= N 13 27 = N^{\frac{13}{27}}

17 Helpful 0 Interesting 0 Brilliant 0 Confused

I want to know more in calculos

Jay Ar GmAn - 7 years, 2 months ago

Log in to reply

Fantastic.

Finn Hulse - 7 years, 1 month ago
Aquilino Madeira
Jul 12, 2015

N N N 3 3 3 = N N 4 3 3 3 = N N 4 9 3 = = N 13 9 3 = N 13 27 = N 13 27 \begin{array}{l} \sqrt[3]{{N\sqrt[3]{{N\sqrt[3]{N}}}}} = \sqrt[3]{{N\sqrt[3]{{\sqrt[3]{{{N^4}}}}}}} = \sqrt[3]{{N\sqrt[9]{{{N^4}}}}} = \\ = \sqrt[3]{{\sqrt[9]{{{N^{13}}}}}} = \sqrt[{27}]{{{N^{13}}}} = {N^{\frac{{13}}{{27}}}} \end{array}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...