Difference between revisions of "2003 AMC 12A Problems/Problem 24"
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<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 <math>2</math>, so the largest possible values is <math>2-2=\ | + | 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=0 \Rightarrow{\textbf{B}}.</math> |
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| + | Note that the maximum occurs when <math>a=b</math>. | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2003|ab=A|num-b=23|num-a=25}} | {{AMC12 box|year=2003|ab=A|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 22:57, 4 January 2019
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 
Note that the maximum occurs when 
.
See Also
| 2003 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 23 |
Followed by Problem 25 |
| 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. 
