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 == |
− | |||
− | |||
First, we have: | First, we have: | ||
Line 53: | Line 51: | ||
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 | + | 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 find |
− | <math>a^{(log_a27)(log_37)} | + | <math>a^{(\log_a27)(\log_37)} + b^{(\log_b49)(\log_711)} + c^{(\log_c\sqrt {11})(\log_{11}25)}</math>. |
− | We know that <math>x^{log_xy} | + | We know that <math>x^{\log_xy} =y</math>, so we find |
− | <math>27^{log_37} | + | <math>27^{\log_37} + 49^{\log_711} + \sqrt {11}^{\log_{11}25}</math> |
− | <math>(3^{log_37})^3 | + | <math>(3^{\log_37})^3 + (7^{\log_711})^2 + ({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 |
− | <math>7^3 | + | <math>7^3 + 11^2 + 25^{1/2}</math> |
− | + | <math>= 343 + 121 + 5</math> | |
− | + | <math>= \boxed {469}</math>. | |
== See Also == | == See Also == | ||
{{AIME box|year=2009|n=II|num-b=1|num-a=3}} | {{AIME box|year=2009|n=II|num-b=1|num-a=3}} | ||
+ | [[Category: Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 20:48, 9 August 2020
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 find
.
We know that , so we find
.
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.