2013 AMC 12B Problems/Problem 22: Difference between revisions
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==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 21:20, 7 August 2023
Problem
Let 
and 
be integers. Suppose that the product of the solutions for 
of the equation
![\[8(\log_n x)(\log_m x)-7\log_n x-6 \log_m x-2013 = 0\]](http://latex.artofproblemsolving.com/1/9/5/19555089c6b0a7647279aa6d4dd6068be044d7e2.png)
is the smallest possible integer. What is 
?
Solution
Rearranging logs, the original equation becomes
![\[\frac{8}{\log n \log m}(\log x)^2 - \left(\frac{7}{\log n}+\frac{6}{\log m}\right)\log x - 2013 = 0\]](http://latex.artofproblemsolving.com/3/d/f/3df6ef64bacddd85e8c6a3ba01b31c94058d9d4d.png)
By Vieta's Theorem, the sum of the possible values of 
is ![$\frac{\frac{7}{\log n}+\frac{6}{\log m}}{\frac{8}{\log n \log m}} = \frac{7\log m + 6 \log n}{8} = \log \sqrt[8]{m^7n^6}$](http://latex.artofproblemsolving.com/1/1/4/114330833b4fd99deddc3ef2acb4bbb881d14a26.png)
. 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 ![$\sqrt[8]{m^7n^6}$](http://latex.artofproblemsolving.com/f/1/a/f1a1a942f13056ed570065e4481a68bc9ecaa395.png)
.
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
