Difference between revisions of "2009 AIME II Problems/Problem 2"
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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 | ||
− | <math>a^{(\log_a27)(\log_37)}</math> + <math>b^{(\log_b49)(\log_711)</math> + <math>c^{(\log_c\sqrt {11})(\log_{11}25)}</math> | + | <math>a^{(\log_a27)(\log_37)}</math> + <math>b^{(\log_b49)(\log_711)}</math> + <math>c^{(\log_c\sqrt {11})(\log_{11}25)}</math> |
We know that <math>x^{\log_xy}</math> = <math>y</math>, so we get | We know that <math>x^{\log_xy}</math> = <math>y</math>, so we get | ||
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<math>27^{\log_37}</math> + <math>49^{\log_711}</math> + <math>\sqrt {11}^{\log_{11}25}</math> | <math>27^{\log_37}</math> + <math>49^{\log_711}</math> + <math>\sqrt {11}^{\log_{11}25}</math> | ||
− | <math>(3^{\log_37})^3</math> + <math>(7^{\log_711})^2</math> + <math>({11^{\log_{11}25})^{1/2}</math> | + | <math>(3^{\log_37})^3</math> + <math>(7^{\log_711})^2</math> + <math>({11^{\log_{11}25}})^{1/2}</math> |
The <math>3</math> and the <math>\log_37</math> cancel out to make <math>7</math>, and we can do this for the other two terms. We obtain | The <math>3</math> and the <math>\log_37</math> cancel out to make <math>7</math>, and we can do this for the other two terms. We obtain |
Revision as of 18:19, 9 March 2015
Contents
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
+ +
We know that = , so we get
+ +
+ +
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.