Difference between revisions of "2000 AIME I Problems/Problem 9"
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\log\frac{2000xy}{2yz}-\log y(\log x-\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 \\ | \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 \\ | \log\frac{x}{z}(1-\log y) &= 0 \\ | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | If <math>1-\log y=0</math> then <math>y=10</math>. Substituting into the first equation yields <math>\log20000=4</math> which is not possible. | ||
+ | |||
+ | If <math>\log\frac{x}{z}=0</math> then <math>\frac{x}{z}=1\Longrightarrow x=z</math>. Substituting into the third equation gets | ||
+ | <cmath>\begin{align*} | ||
+ | \log x^2-(\log x)(\log x) &= 0 \\ | ||
+ | \log x^2-\log x^x &= 0 \\ | ||
+ | \log x^{2-x} &= 0 \\ | ||
+ | x^{2-x} &= 1 \\ | ||
+ | \end{align*}</cmath> | ||
+ | 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.) | ||
+ | |||
+ | Substituting <math>x=1</math> and <math>x=2</math> into the first equation will obtain <math>y=5</math> and <math>y=20</math>, respectively. Thus <math>y_1+y_2=\boxed{025}</math>. | ||
+ | |||
+ | ~ Nafer | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | Let <math>a = \log x</math>, <math>b = \log y</math> and <math>c = \log z</math>. Then the given equations become: | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | \log 2 + a + b - ab = 1 \\ | ||
+ | \log 2 + b + c - bc = 1 \\ | ||
+ | a+c = ac \\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Equating the first and second equations, solving, and factoring, we get <math>a(1-b) = c(1-b) \implies{a = c}</math>. Plugging this result into the third equation, we get <math>c = 0</math> or <math>2</math>. Substituting each of these values of <math>c</math> into the second equation, we get <math>b = 1 - \log 2</math> and <math>b = 1 + \log 2</math>. Substituting backwards from our original substitution, we get <math>y = 5</math> and <math>y = 20</math>, respectively, so our answer is <math>\boxed{025}</math>. | ||
+ | |||
+ | ~ anellipticcurveoverq | ||
+ | |||
+ | ==Video solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=sOyLnGJjVvc&t | ||
== See also == | == See also == |
Revision as of 23:03, 28 January 2021
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 of into the second equation, we get and . Substituting backwards from our original substitution, we get and , respectively, so our answer is .
~ anellipticcurveoverq
Video solution
https://www.youtube.com/watch?v=sOyLnGJjVvc&t
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.