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