Difference between revisions of "2002 AIME I Problems/Problem 6"

Revision as of 10:18, 20 April 2008

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 $\log_{30}\left(x_1y_1x_2y_2\right)$ .

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@@ Line 1: / Line 1: @@
 == Problem ==
 The solutions to the system of equations
+<center><math>\log_{225}x+\log_{64}y=4</math></center>
-<cmath>\log_{225}x+\log_{64}y=4</cmath>
+<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>.
-<cmath>\log_{x}225-\log_{y}64=1</cmath>
-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 ==

2002 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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All AIME Problems and Solutions