Difference between revisions of "2003 AMC 12A Problems/Problem 24"
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Note that the maximum occurs when <math>a=b</math>. | Note that the maximum occurs when <math>a=b</math>. | ||
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+ | ==Solution 2 (More Algebraic)== | ||
+ | Similar to the previous solution, we use our logarithmic rules and start off the same way. | ||
+ | <cmath>\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)</cmath> | ||
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+ | However, now, we use a different logarithmic rule stating that <math>\log_{a}b</math> is simply equal to <math>\frac {\log_{10}b}{\log_{10}a}</math>. With this, we can rewrite our previous equation to give us <cmath>2-(\log_{a}b+\log_{b}a) = 2 - (\frac {\log_{10}b}{\log_{10}a} + \frac {\log_{10}a}{\log_{10}b})</cmath>. | ||
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+ | We can now cross multiply to get that <cmath>2 - (\frac {\log_{10}b}{\log_{10}a} + \frac {\log_{10}a}{\log_{10}b}) = 2 - (\frac {2 * log_{10}b * \log_{10}a}{log_{10}b * \log_{10}a})</cmath> Finally, we cancel to get <math>2-2=0 \Rightarrow \boxed{\textbf{(B) 0}}.</math> | ||
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+ | Solution by: armang32324 | ||
==Video Solution== | ==Video Solution== |
Revision as of 19:35, 6 April 2023
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
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.