Difference between revisions of "2000 AIME I Problems/Problem 9"
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Thus either <math>x=1</math> or <math>2-x=0\Longrightarrow x=2</math>. (Note that here <math>x\neq-1</math> since logarithm isn't defined for negative number.) | Thus either <math>x=1</math> or <math>2-x=0\Longrightarrow x=2</math>. (Note that here <math>x\neq-1</math> since logarithm isn't defined for negative number.) | ||
− | If | + | Going back to the first equation. If <math>x=1</math> then |
== See also == | == See also == |
Revision as of 23:37, 6 December 2019
Contents
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 third equation gets
Thus either
or
. (Note that here
since logarithm isn't defined for negative number.)
Going back to the first equation. If then
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.