Difference between revisions of "2000 AIME I Problems/Problem 9"
(→Solution 2) |
(→Solution 2) |
||
Line 38: | Line 38: | ||
\log1000+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &= 3 \\ | \log1000+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &= 3 \\ | ||
\3+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &= 3 \\ | \3+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &= 3 \\ | ||
− | + | \log\frac{x}{z}(1-\log y) &= 0 \\ | |
\end{align*}</cmath> | \end{align*}</cmath> | ||
Revision as of 23:28, 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
\begin{align*} \log 2000xy-\log 2yz-((\log x)(\log y)-(\log y)(\log z)) &= 3 \\ \log\frac{2000xy}{2yz}-\log y(\log x-\log z) &= 3 \\ \log1000+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &= 3 \\ \3+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &= 3 \\ \log\frac{x}{z}(1-\log y) &= 0 \\ \end{align*} (Error compiling LaTeX. Unknown error_msg)
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.