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| − | ==Problem==
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| − | Let <math>m>1</math> and <math>n>1</math> be integers. Suppose that the product of the solutions for <math>x</math> of the equation
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| − | <cmath> 8(\log_n x)(\log_m x)-7\log_n x-6 \log_m x-2013 = 0 </cmath>
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| − | is the smallest possible integer. What is <math>m+n</math>?
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| − | <math> \textbf{(A)}\ 12\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 272 </math>
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| − | ==Solution==
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| − | Rearranging logs, the original equation becomes
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| − | <cmath>\frac{8}{\log n \log m}(\log x)^2 - \left(\frac{7}{\log n}+\frac{6}{\log m}\right)\log x - 2013 = 0</cmath>
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| − | By Vieta's Theorem, the sum of the possible values of <math>\log x</math> is <math>\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}</math>. But the sum of the possible values of <math>\log x</math> is the logarithm of the product of the possible values of <math>x</math>. Thus the product of the possible values of <math>x</math> is equal to <math>\sqrt[8]{m^7n^6}</math>.
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| − | It remains to minimize the integer value of <math>\sqrt[8]{m^7n^6}</math>. Since <math>m, n>1</math>, we can check that <math>m = 2^2</math> and <math>n = 2^3</math> work. Thus the answer is <math>4+8 = \boxed{\textbf{(A)}\ 12}</math>.
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| − | ==Video Solution==
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| − | For those who prefer a video solution: https://www.youtube.com/watch?v=vX0y9lRv9OM&t=312s
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| − | == See also ==
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| − | {{AMC12 box|year=2013|ab=B|num-b=21|num-a=23}}
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| − | {{MAA Notice}}
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