2009 AIME II Problems/Problem 2
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
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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 |
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