2002 AMC 12P Problems/Problem 21: Difference between revisions
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Factoring gives <math>(2x-y)(x-2y) = 0</math> or <math>\frac {x}{y} = 2, \frac {1}{2}</math>. | Factoring gives <math>(2x-y)(x-2y) = 0</math> or <math>\frac {x}{y} = 2, \frac {1}{2}</math>. | ||
Recall that <math>\frac {x}{y} = \frac {\log a}{\log b} = \log_{a} b</math>. Therefore, the maximum value of <math>\log_{a} b</math> is | Recall that <math>\frac {x}{y} = \frac {\log a}{\log b} = \log_{a} b</math>. Therefore, the maximum value of <math>\log_{a} b</math> is <math>\boxed {text{(C) }2}</math>. | ||
== See also == | == See also == | ||
{{AMC12 box|year=2002|ab=P|num-b=20|num-a=22}} | {{AMC12 box|year=2002|ab=P|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:02, 10 March 2024
Problem
Let 
and 
be real numbers greater than 
for which there exists a positive real number 
different from , such that
![\[2(\log_a{c} + \log_b{c}) = 9\log_{ab}{c}.\]](http://latex.artofproblemsolving.com/d/3/c/d3cf084efbe6ea20e4571491e473bcc3aa15c877.png)
Find the largest possible value of 
Solution
We may rewrite the given equation as ![\[2(\frac {\log c}{\log a} + \frac {\log c}{\log b}) = \frac {9\log c}{\log a + \log b}\]](http://latex.artofproblemsolving.com/d/4/f/d4f14abfdce234e593a493f6c1c59dbd269bf112.png)
Since 
, we have 
, so we may divide by 
on both sides. After making the substitutions 
and 
, our equation becomes ![\[\frac {2}{x} + \frac {2}{y} = \frac {9}{x+y}\]](http://latex.artofproblemsolving.com/e/8/4/e84b7e9daac34552d84eeb2e3a391d4ba9b35a00.png)
Rewriting the left-hand side gives ![\[\frac {2(x+y)}{xy} = \frac {9}{x+y}\]](http://latex.artofproblemsolving.com/1/7/7/1779e615a828581c035db9233bcbaadb58c954d2.png)
Cross-multiplying gives 
or ![\[2x^2 - 5xy + 2y^2 = 0\]](http://latex.artofproblemsolving.com/2/e/a/2eac863e98398cc2843c64f77363c8f31d9abd89.png)
Factoring gives 
or 
.
Recall that 
. Therefore, the maximum value of 
is 
.
See also
| 2002 AMC 12P (Problems • Answer Key • Resources) | |
| Preceded by Problem 20 |
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 | |
These problems are copyrighted © by the Mathematical Association of America. 
