Difference between revisions of "2013 AMC 12B Problems/Problem 22"
Line 2: | Line 2: | ||
==Problem== | ==Problem== | ||
− | + | ||
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 | 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 | ||
− | + | ||
<cmath> 8(\log_n x)(\log_m x)-7\log_n x-6 \log_m x-2013 = 0 </cmath> | <cmath> 8(\log_n x)(\log_m x)-7\log_n x-6 \log_m x-2013 = 0 </cmath> | ||
− | + | ||
is the smallest possible integer. What is <math>m+n</math>? | is the smallest possible integer. What is <math>m+n</math>? | ||
− | + | ||
<math> \textbf{(A)}\ 12\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 272 </math> | <math> \textbf{(A)}\ 12\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 272 </math> | ||
− | + | ||
− | + | ||
==Solution== | ==Solution== | ||
− | + | ||
Rearranging logs, the original equation becomes | Rearranging logs, the original equation becomes | ||
− | + | ||
<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> | <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> | ||
− | + | ||
− | + | ||
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>. | ||
− | + | ||
− | + | ||
− | + | ||
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>. | 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>. | ||
− | + | ||
− | + | ||
==Video Solution== | ==Video Solution== | ||
− | + | ||
For those who prefer a video solution: https://www.youtube.com/watch?v=vX0y9lRv9OM&t=312s | For those who prefer a video solution: https://www.youtube.com/watch?v=vX0y9lRv9OM&t=312s | ||
− | + | ||
− | + | ||
== 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 22:20, 7 August 2023
Contents
Problem
Let and
be integers. Suppose that the product of the solutions for
of the equation
is the smallest possible integer. What is ?
Solution
Rearranging logs, the original equation becomes
By Vieta's Theorem, the sum of the possible values of is
. But the sum of the possible values of
is the logarithm of the product of the possible values of
. Thus the product of the possible values of
is equal to
.
It remains to minimize the integer value of . Since
, we can check that
and
work. Thus the answer is
.
Video Solution
For those who prefer a video solution: https://www.youtube.com/watch?v=vX0y9lRv9OM&t=312s
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.