Difference between revisions of "2009 AIME II Problems/Problem 2"
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<cmath> a^{(\log_3 7)^2} + b^{(\log_7 11)^2} + c^{(\log_{11} 25)^2}. </cmath> | <cmath> a^{(\log_3 7)^2} + b^{(\log_7 11)^2} + c^{(\log_{11} 25)^2}. </cmath> | ||
− | == Solution == | + | == Solution 1 == |
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First, we have: | First, we have: | ||
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and therefore the answer is <math>343+121+5 = \boxed{469}</math>. | and therefore the answer is <math>343+121+5 = \boxed{469}</math>. | ||
− | + | == Solution 2 == | |
We know from the first three equations that <math>\log_a27</math> = <math>\log_37</math>, <math>\log_b49</math> = <math>\log_711</math>, and <math>\log_c\sqrt{11}</math> = <math>\log_{11}25</math>. Substituting, we get | We know from the first three equations that <math>\log_a27</math> = <math>\log_37</math>, <math>\log_b49</math> = <math>\log_711</math>, and <math>\log_c\sqrt{11}</math> = <math>\log_{11}25</math>. Substituting, we get |
Revision as of 14:49, 8 March 2014
Contents
[hide]Problem
Suppose that , , and are positive real numbers such that , , and . Find
Solution 1
First, we have:
Now, let , then we have:
This is all we need to evaluate the given formula. Note that in our case we have , , and . We can now compute:
Similarly, we get
and
and therefore the answer is .
Solution 2
We know from the first three equations that = , = , and = . Substituting, we get
+ $b^{(\log_b49)(\log_711)$ (Error compiling LaTeX. Unknown error_msg) +
We know that = , so we get
+ +
+ + $({11^{\log_{11}25})^{1/2}$ (Error compiling LaTeX. Unknown error_msg)
The and the cancel out to make , and we can do this for the other two terms. We obtain
+ +
= + + = .
See Also
2009 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.