Difference between revisions of "2003 AMC 12A Problems/Problem 24"
(→See Also) |
(→See Also) |
||
| Line 22: | Line 22: | ||
== See Also == | == See Also == | ||
| − | {{AMC12 box|year=2003|ab=A|num-b=23|after=[ | + | {{AMC12 box|year=2003|ab=A|num-b=23|after=[http://www.artofproblemsolving.com/Wiki/index.php/2003_AMC_12A_Problems/Problem_25]}} |
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 19:10, 15 January 2014
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/a/c/8/ac81a63197833883d0f23b2c94e96af4303e9c3e.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 23 |
Followed by [1] |
| 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. 
