Difference between revisions of "2002 AIME I Problems/Problem 6"
m (→See also) |
m (→Solution) |
||
(5 intermediate revisions by 5 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
The solutions to the system of equations | The solutions to the system of equations | ||
+ | <center><math>\log_{225}x+\log_{64}y=4</math></center> | ||
+ | <center><math>\log_{x}225-\log_{y}64=1</math></center> | ||
+ | are <math>(x_1,y_1)</math> and <math>(x_2,y_2)</math>. Find <math>\log_{30}\left(x_1y_1x_2y_2\right)</math>. | ||
− | < | + | == Solution == |
+ | Let <math>A=\log_{225}x</math> and let <math>B=\log_{64}y</math>. | ||
+ | |||
+ | From the first equation: <math>A+B=4 \Rightarrow B = 4-A</math>. | ||
+ | |||
+ | Plugging this into the second equation yields <math>\frac{1}{A}-\frac{1}{B}=\frac{1}{A}-\frac{1}{4-A}=1 \Rightarrow A = 3\pm\sqrt{5}</math> and thus, <math>B=1\pm\sqrt{5}</math>. | ||
− | < | + | So, <math>\log_{225}(x_1x_2)=\log_{225}(x_1)+\log_{225}(x_2)=(3+\sqrt{5})+(3-\sqrt{5})=6</math> <math>\Rightarrow x_1x_2=225^6=15^{12}</math>. |
− | + | And <math>\log_{64}(y_1y_2)=\log_{64}(y_1)+\log_{64}(y_2)=(1+\sqrt{5})+(1-\sqrt{5})=2</math> <math>\Rightarrow y_1y_2=64^2=2^{12}</math>. | |
− | == | + | Thus, <math>\log_{30}\left(x_1y_1x_2y_2\right) = \log_{30}\left(15^{12}\cdot2^{12} \right) = \log_{30}\left(30^{12} \right) = \boxed{012}</math>. |
− | {{ | + | |
+ | |||
+ | One may simplify the work by applying [[Vieta's formulas]] to directly find that <math>\log x_1 + \log x_2 = 6 \log 225, \log y_1 + \log y_2 = 2 \log 64</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2002|n=I|num-b=5|num-a=7}} | {{AIME box|year=2002|n=I|num-b=5|num-a=7}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 18:55, 23 April 2016
Problem
The solutions to the system of equations
are and . Find .
Solution
Let and let .
From the first equation: .
Plugging this into the second equation yields and thus, .
So, .
And .
Thus, .
One may simplify the work by applying Vieta's formulas to directly find that .
See also
2002 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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.