# Difference between revisions of "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 $m+n$?

$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 24\qquad\textbf{(D}}\ 48\qquad\textbf{(E)}\ 272$ (Error compiling LaTeX. ! Extra }, or forgotten \$.)

## 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 $\frac{\frac{7}{\log n}+\frac{6}{\log m}}{\frac{8}{\log n \log m}} = \frac{7\log m + 6 \log n}{8} = \log \sqrt{m^7n^6}$. 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 $\sqrt{m^7n^6}$.

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

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