Logarithmic Equation Product
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 $?
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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{12} $.
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