2000 AIME I Problems/Problem 9
The system of equations
has two solutions and . Find .
Since , we can reduce the equations to a more recognizable form:
Let be respectively. Using SFFT, the above equations become (*)
Small note from different author:
From here, multiplying the three equations gives
Dividing the third equation of (*) from this equation, . (Note from different author if you are confused on this step: if is positive then so if is negative then so ) This gives , and the answer is .
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 .
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 of into the second equation, we get and . Substituting backwards from our original substitution, we get and , respectively, so our answer is .
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