Difference between revisions of "2000 AIME I Problems/Problem 9"
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(a-1)(c-1) &=& 1 | (a-1)(c-1) &=& 1 | ||
\end{eqnarray*}</cmath> | \end{eqnarray*}</cmath> | ||
+ | |||
+ | Small note from different author: <math>-(3 - \log 2000) = \log 2000 - 3 = \log 2000 - \log 1000 = \log 2.</math> | ||
From here, multiplying the three equations gives | From here, multiplying the three equations gives | ||
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(a-1)(b-1)(c-1) &=& \pm\log 2\end{eqnarray*}</cmath> | (a-1)(b-1)(c-1) &=& \pm\log 2\end{eqnarray*}</cmath> | ||
− | Dividing the third equation of (*) from this equation, <math>b-1 = \log y - 1 = \pm\log 2 \Longrightarrow \log y = \pm \log 2 + 1</math>. This gives <math>y_1 = 20, y_2 = 5</math>, and the answer is <math>y_1 + y_2 = \boxed{025}</math>. | + | Dividing the third equation of (*) from this equation, <math>b-1 = \log y - 1 = \pm\log 2 \Longrightarrow \log y = \pm \log 2 + 1</math>. (Note from different author if you are confused on this step: if <math>\pm</math> is positive then <math>\log y = \log 2 + 1 = \log 2 + \log 10 = \log 20,</math> so <math>y=20.</math> if <math>\pm</math> is negative then <math>\log y = 1 - \log 2 = \log 10 - \log 2 = \log 5,</math> so <math>y=5.</math>) This gives <math>y_1 = 20, y_2 = 5</math>, and the answer is <math>y_1 + y_2 = \boxed{025}</math>. |
== Solution 2 == | == Solution 2 == | ||
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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.) | ||
− | 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{ | + | 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 | ~ 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: | Let <math>a = \log x</math>, <math>b = \log y</math> and <math>c = \log z</math>. Then the given equations become: | ||
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\end{align*}</cmath> | \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 - \ | + | 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 | ~ anellipticcurveoverq | ||
+ | |||
+ | ==Video solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=sOyLnGJjVvc&t | ||
== See also == | == See also == |
Latest revision as of 15:56, 8 May 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 (*)
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 .
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 |
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