Difference between revisions of "2003 AMC 12A Problems/Problem 24"
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The Link: https://www.youtube.com/watch?v=InF2phZZi2A&t=1s | The Link: https://www.youtube.com/watch?v=InF2phZZi2A&t=1s | ||
| + | -MistyMathMusic | ||
== See Also == | == See Also == | ||
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Revision as of 22:36, 18 August 2020
Contents
Problem
If
what is the largest possible value of
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
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.