Difference between revisions of "2003 AMC 12A Problems/Problem 23"
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Since <math>a</math> and <math>b</math> are both positive, using [[AM-GM]] gives that the term in parentheses must be at least <math>2</math>, so the largest possible values is <math>2-2=\boxed{0}.</math> | Since <math>a</math> and <math>b</math> are both positive, using [[AM-GM]] gives that the term in parentheses must be at least <math>2</math>, so the largest possible values is <math>2-2=\boxed{0}.</math> | ||
| + | |||
| + | == See Also == | ||
| + | {{AMC12 box|year=2003|ab=A|num-b=22|num-a=24}} | ||
| + | |||
| + | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 17:36, 22 February 2010
Problem
If 
what is the largest possible value of 
Solution
Using logarithmic rules, we see that
![\[\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a\]](http://latex.artofproblemsolving.com/9/e/9/9e95d3760b7b2da7f5478020c500dc7d4e724758.png)
![\[=2-(\log_{a}b+\frac {1}{\log_{a}b})\]](http://latex.artofproblemsolving.com/8/6/9/86936dda057f6fa26fa792fcec247eb7fd350936.png)
Since 
and 
are both positive, using AM-GM gives that the term in parentheses must be at least 
, so the largest possible values is 
See Also
| 2003 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 22 |
Followed by Problem 24 |
| 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 | |
