Difference between revisions of "2009 AIME II Problems/Problem 2"
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Idk12345678 (talk | contribs) |
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<cmath>= 343 + 121 + 5</cmath> | <cmath>= 343 + 121 + 5</cmath> | ||
<cmath>= \boxed {469}.</cmath> | <cmath>= \boxed {469}.</cmath> | ||
+ | |||
+ | == Solution 3 == | ||
+ | First, let us take the log base 3 of the first expression. We get <math>\log_3{a^{\log_3{7}}} = 3</math>. Simplifying, we get <cmath>(\log_3{7})(\log_3{a}) = 3</cmath>. So, <cmath>\log_3 a = \frac{3}{\log_3{7}}</cmath>, and <cmath>a = 3^\frac{3}{log_3{7}}</cmath>. We can repeat the same process for the other equations, giving us <cmath>b = 7^\frac{2}{\log_7{11}}</cmath>, and <cmath>c = (\sqrt{11})^\frac{1}{\log_ | ||
+ | {11}{25}}</cmath>. Raising <math>a</math> to the power of <math>(\log_3{7})^2</math>, we get <cmath>3^{3\log_3{7}} = 3^{\log_3{343}} = 343</cmath>. Repeating a similar process for the other ones(you have to turn the square root to a fractional power for c), we get <math>343+121+5 = \boxed{469}</math> | ||
+ | |||
+ | ~idk12345678 | ||
== See Also == | == See Also == |
Latest revision as of 14:20, 10 May 2024
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 find
We know that , so we find
The and the cancel to make , and we can do this for the other two terms. Thus, our answer is
Solution 3
First, let us take the log base 3 of the first expression. We get . Simplifying, we get . So, , and . We can repeat the same process for the other equations, giving us , and . Raising to the power of , we get . Repeating a similar process for the other ones(you have to turn the square root to a fractional power for c), we get
~idk12345678
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.