Difference between revisions of "2002 AMC 12P Problems/Problem 21"
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== Solution == | == Solution == | ||
We may rewrite the given equation as <math>2(\frac {\log c}{\log a} + \frac {\log c}{\log b}) = \frac {9\log c}{\log a + \log b}</math>. | We may rewrite the given equation as <math>2(\frac {\log c}{\log a} + \frac {\log c}{\log b}) = \frac {9\log c}{\log a + \log b}</math>. | ||
| − | Since <math>c \neq 1</math> | + | Since <math>c \neq 1</math>, we have <math>\log c \neq 0</math>, so we may divide by <math>\log c</math> on both sides. After making the substitutions <math>x = \log a</math> and <math>y = \log b</math>, our equation becomes |
== 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 14:53, 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 
.
Since 
, we have 
, so we may divide by 
on both sides. After making the substitutions 
and 
, our equation becomes
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
