Difference between revisions of "2000 AIME I Problems/Problem 9"
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~ Nafer | ~ 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 for <math>c</math> into the second equation, we get <math>b = 1 - \log_{10}2</math> and <math>b = 1 + \log_{10}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 | ||
== See also == | == See also == |
Revision as of 17:16, 9 July 2020
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.