Difference between revisions of "2003 AMC 12A Problems/Problem 23"

(Solution)
(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>
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<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 two, 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>

Revision as of 18:33, 22 February 2010

Solution

Using logarithmic rules, we see that

\[\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a\] \[=2-(\log_{a}b+\frac {1}{\log_{a}b})\]

Since $a$ and $b$ are both positive, using AM-GM gives that the term in parentheses must be at least $2$, so the largest possible values is $2-2=\boxed{0}.$