Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2013 AMC 12B Problems/Problem 22"
 
Line 43: Line 43:
 
 
 
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
−
+
 
−
+
==Video Solution 2 by MOP 2024==
−
+
https://youtu.be/n5RfHdh3HTk
 +
 
 +
~r00tsOfUnity
 
 
 
== See also ==
 
== See also ==
Line 52: Line 54:
 
{{AMC12 box|year=2013|ab=B|num-b=21|num-a=23}}
 
{{AMC12 box|year=2013|ab=B|num-b=21|num-a=23}}
 
   
 
   
−
+
[[Category:Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 01:17, 26 August 2023

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

Video Solution 2 by MOP 2024

https://youtu.be/n5RfHdh3HTk

~r00tsOfUnity

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_22&oldid=197488"