2003 AMC 12A Problems/Problem 24
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 .
Solution 2 (More Algebraic)
Similar to the previous solution, we use our logarithmic rules and start off the same way.
However, now, we use a different logarithmic rule stating that is simply equal to . With this, we can rewrite our previous equation to give us .
We can now cross multiply to get that Finally, we cancel to get
Solution by: armang32324
Solution 3 (Calculus)
By the logarithmic rules, we have . Let . Thus, the expression becomes . We want to find the maximum of the function. To do so, we will find its derivative and let it equal to 0. We get: Since is not a solution. Thus, . Substituting it into the original expression , we get .
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 |
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