2000 AIME I Problems/Problem 9
Problem
The system of equations
has two solutions and . Find .
Solution
Since , we can reduce the equations to a more recognizable form:
Let be respectively. Using SFFT, the above equations become (*)
From here, multiplying the three equations gives
Dividing the third equation of (*) from this equation, . This gives , and the answer is .
Solution 2
Subtracting the second equation from the first equation yields If then . Substituting into the first equation yields which is not possible.
If then . Substituting into the third equation gets Thus either or . (Note that here since logarithm isn't defined for negative number.)
Substituting and into the first equation will obtain and , respectively. Thus .
~ Nafer
Solution 3
Let , and . Then the given equations become:
Equating the first and second equations, solving, and factoring, we get . Plugging this result into the third equation, we get or . Substituting each of these values for into the second equation, we get and . Substituting backwards from our original substitution, we get and , respectively, so our answer is .
~ anellipticcurveoverq
See also
2000 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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