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 | + | Since <math>a</math> and <math>b</math> are both greater than <math>1</math>, 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 \boxed{\textbf{B}}.</math> |
Note that the maximum occurs when <math>a=b</math>. | Note that the maximum occurs when <math>a=b</math>. |
Latest revision as of 12:03, 19 April 2021
Contents
Problem
If what is the largest possible value of
Solution
Using logarithmic rules, we see that
Since and are both greater than , 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 .
Video Solution
The Link: https://www.youtube.com/watch?v=InF2phZZi2A&t=1s
-MistyMathMusic
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.