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

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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>.
 
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 = 2^3</math> and <math>n = 2^2</math> work. Thus the answer is <math>4+8 = \boxed{\textbf{(A)}\ 12}</math>.
+
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>.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2013|ab=B|num-b=21|num-a=23}}
 
{{AMC12 box|year=2013|ab=B|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:26, 1 January 2016

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 $4+8 = \boxed{\textbf{(A)}\ 12}$.

See also

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

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