2002 AIME I Problems/Problem 6: Difference between revisions

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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 $\log x_1 + \log x_2 = 6 \log 225, \log y_1 + \log y_2 = 2 \log 64$ .

@@ Line 17: / Line 17: @@
 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 ==
 {{AIME box|year=2002|n=I|num-b=5|num-a=7}}
+[[Category:Intermediate Algebra Problems]]
+{{MAA Notice}}

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2002 AIME I Problems/Problem 6: Difference between revisions

Latest revision as of 18:55, 23 April 2016

Problem

Solution

See also