Difference between revisions of "Mock AIME I 2015 Problems/Problem 2"

Revision as of 20:55, 10 June 2015

Use the logarithmic identity $\log_q p = \frac{log_r p}{log_r q}$ to expand the assumptions to

$\log_x 3y = \frac{log_3 3y}{log_3 x} = \frac{1+log_3 y}{log_3 x} = \frac{20}{13}$

and

$\log_{3x} y = \frac{log_3 y}{log_3 3x} = \frac{log_3 y}{1+log_3 x} = \frac{2}{3}.$

Solve for the values of $\log_3 x$ and $\log_3 y$ which are respectively $\frac{65}{34}$ and $\frac{33}{17}.$

The sought ratio is

$\log_{3x} 3y = \frac{log_3 3y}{log_3 3x} = \frac{1+log_3 y}{1+log_3 x} = \frac{1+\tfrac{33}{17}}{1+\tfrac{65}{34}} = \frac{100}{99}.$

The answer then is $100+99=199.$

Solution by D. Adrian Tanner (Original solution and answer below - origin unknown)

By rearranging the values, it is possible to attain an

x= 3^ (65/17)

and

y= 3^ (33/17)

Therefore, a/b is equal to 25/61, so 25+41= 066

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+==Corrected Solution and Answer==
+Use the logarithmic identity  <math>\log_q p = \frac{log_r p}{log_r q}</math> to expand the assumptions to
+<math>\log_x 3y = \frac{log_3 3y}{log_3 x}  = \frac{1+log_3 y}{log_3 x} = \frac{20}{13}</math>
+and
+<math>\log_{3x} y = \frac{log_3 y}{log_3 3x}  = \frac{log_3 y}{1+log_3 x} = \frac{2}{3}.</math>
+Solve for the values of <math>\log_3 x</math> and <math>\log_3 y</math> which are respectively <math> \frac{65}{34}</math> and <math>\frac{33}{17}.</math>
+The sought ratio is
+<math>\log_{3x} 3y = \frac{log_3 3y}{log_3 3x}  = \frac{1+log_3 y}{1+log_3 x} = \frac{1+\tfrac{33}{17}}{1+\tfrac{65}{34}} = \frac{100}{99}.</math>
+The answer then is <math>100+99=199.</math>
+Solution by D. Adrian Tanner
+(Original solution and answer below - origin unknown)
+***********************************************************************
 By rearranging the values, it is possible to attain an