Difference between revisions of "2013 AMC 12B Problems/Problem 22"
Turkeybob777 (talk | contribs) |
|||
Line 1: | Line 1: | ||
==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>? | ||
Line 13: | Line 13: | ||
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>. | ||
+ | |||
+ | == See also == | ||
+ | {{AMC12 box|year=2013|ab=B|num-b=21|num-a=22}} |
Revision as of 17:15, 22 February 2013
Problem
Let and be integers. Suppose that the product of the solutions for of the equation is the smallest possible integer. What is ?
$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 24\qquad\textbf{(D}}\ 48\qquad\textbf{(E)}\ 272$ (Error compiling LaTeX. Unknown error_msg)
Solution
Rearranging logs, the original equation becomes
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 .
See also
2013 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 22 |
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 |