Difference between revisions of "2003 AMC 12A Problems/Problem 23"
Fuzzy growl (talk | contribs) (Created page with '== Solution == Using logarithmic rules, we see that <cmath>\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a</cmath> <math>=2-(\log_{a}b+\frac {1}{\log_{a}b}</ma…') |
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<cmath>\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a</cmath> | <cmath>\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a</cmath> | ||
| − | < | + | <cmath>=2-(\log_{a}b+\frac {1}{\log_{a}b}</cmath> |
| − | Since < | + | Since <math>a</math> and <math>b</math> are both positive, using [[AM-GM]] gives that the term in parentheses must be at least two, so the largest possible values is <math>2-2=\boxed{0}.</math> |
Revision as of 18:33, 22 February 2010
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/7/1/9/7199c65b4d4a6c8e9132bf881ef3d3dd77315fad.png)
Since 
and 
are both positive, using AM-GM gives that the term in parentheses must be at least two, so the largest possible values is 
