Difference between revisions of "2013 AMC 12B Problems/Problem 22"

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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}}

Revision as of 11:09, 7 December 2021

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

See also

2013 AMC 12B (ProblemsAnswer KeyResources)
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

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