2002 AIME I Problems/Problem 6: Difference between revisions
I like pie (talk | contribs) mNo edit summary |
|||
| (4 intermediate revisions by 4 users not shown) | |||
| Line 6: | Line 6: | ||
== Solution == | == 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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
