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Difference between revisions of "2002 AIME I Problems/Problem 6"
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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>.  
 
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>.  
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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}}
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[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 18:55, 23 April 2016

Problem

The solutions to the system of equations

$\log_{225}x+\log_{64}y=4$
$\log_{x}225-\log_{y}64=1$

are $(x_1,y_1)$ and $(x_2,y_2)$. Find $\log_{30}\left(x_1y_1x_2y_2\right)$.

Solution

Let $A=\log_{225}x$ and let $B=\log_{64}y$.

From the first equation: $A+B=4 \Rightarrow B = 4-A$.

Plugging this into the second equation yields $\frac{1}{A}-\frac{1}{B}=\frac{1}{A}-\frac{1}{4-A}=1 \Rightarrow A = 3\pm\sqrt{5}$ and thus, $B=1\pm\sqrt{5}$.

So, $\log_{225}(x_1x_2)=\log_{225}(x_1)+\log_{225}(x_2)=(3+\sqrt{5})+(3-\sqrt{5})=6$ $\Rightarrow x_1x_2=225^6=15^{12}$.

And $\log_{64}(y_1y_2)=\log_{64}(y_1)+\log_{64}(y_2)=(1+\sqrt{5})+(1-\sqrt{5})=2$ $\Rightarrow y_1y_2=64^2=2^{12}$.

Thus, $\log_{30}\left(x_1y_1x_2y_2\right) = \log_{30}\left(15^{12}\cdot2^{12} \right) = \log_{30}\left(30^{12} \right) = \boxed{012}$.


One may simplify the work by applying Vieta's formulas to directly find that $\log x_1 + \log x_2 = 6 \log 225, \log y_1 + \log y_2 = 2 \log 64$.

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

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