Difference between revisions of "1995 AIME Problems/Problem 2"

(hastily written solution, box)
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== Problem ==
 
== Problem ==
Find the last three digits of the product of the positive roots of
+
Find the last three digits of the product of the [[positive root]]s of
<math>\sqrt{1995}x^{\log_{1995}x}=x^2.</math>
+
<math>\sqrt{1995}x^{\log_{1995}x}=x^2</math>.
  
 
== Solution ==
 
== Solution ==
 +
Taking the <math>\log_{1995}</math> ([[logarithm]]) of both sides and then moving to one side yields a [[quadratic equation]]: <math>2(\log_{1995}x)^2 - 4(\log_{1995}x)  + 1 = 0</math>. Applying the [[quadratic formula]] yields that <math>\log_{1995}x = \frac{2 \pm \sqrt{2}}{2}</math>. Thus, the product of the two roots is <math>= 1995^2</math>, making the solution <math>025</math>.
  
 
== See also ==
 
== See also ==
* [[1995_AIME_Problems/Problem_1|Previous Problem]]
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{{AIME box|year=1995|num-b=1|num-a=3}}
* [[1995_AIME_Problems/Problem_3|Next Problem]]
 
* [[1995 AIME Problems]]
 

Revision as of 22:29, 13 February 2007

Problem

Find the last three digits of the product of the positive roots of $\sqrt{1995}x^{\log_{1995}x}=x^2$.

Solution

Taking the $\log_{1995}$ (logarithm) of both sides and then moving to one side yields a quadratic equation: $2(\log_{1995}x)^2 - 4(\log_{1995}x)  + 1 = 0$. Applying the quadratic formula yields that $\log_{1995}x = \frac{2 \pm \sqrt{2}}{2}$. Thus, the product of the two roots is $= 1995^2$, making the solution $025$.

See also

1995 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions