2013 AMC 12B Problems/Problem 22

Problem

− Let $m>1$ and $n>1$ be integers. Suppose that the product of the solutions for $x$ of the equation

− $8(\log_n x)(\log_m x)-7\log_n x-6 \log_m x-2013 = 0$

− is the smallest possible integer. What is $m+n$ ?

− $\textbf{(A)}\ 12\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 272$

−

Solution

− Rearranging logs, the original equation becomes

− $\frac{8}{\log n \log m}(\log x)^2 - \left(\frac{7}{\log n}+\frac{6}{\log m}\right)\log x - 2013 = 0$

−

− By Vieta's Theorem, the sum of the possible values of $\log x$ is $\frac{\frac{7}{\log n}+\frac{6}{\log m}}{\frac{8}{\log n \log m}} = \frac{7\log m + 6 \log n}{8} = \log \sqrt[8]{m^7n^6}$ . But the sum of the possible values of $\log x$ is the logarithm of the product of the possible values of $x$ . Thus the product of the possible values of $x$ is equal to $\sqrt[8]{m^7n^6}$ .

−

− It remains to minimize the integer value of $\sqrt[8]{m^7n^6}$ . Since $m, n>1$ , we can check that $m = 2^2$ and $n = 2^3$ work. Thus the answer is $4+8 = \boxed{\textbf{(A)}\ 12}$ .

−

Video Solution

− For those who prefer a video solution: https://www.youtube.com/watch?v=vX0y9lRv9OM&t=312s

−

2013 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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
All AMC 12 Problems and Solutions

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

Page

Toolbox

Search

2013 AMC 12B Problems/Problem 22

Contents

Problem

Solution

Video Solution

See also