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

Revision as of 13:04, 3 July 2013

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. Unknown error_msg)

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

2013 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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All AMC 12 Problems and Solutions

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

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 == See also ==
 {{AMC12 box|year=2013|ab=B|num-b=21|num-a=23}}
+{{MAA Notice}}

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

Revision as of 13:04, 3 July 2013

Problem

Solution

See also