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

(Incorrect solution and answer resolved)
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==Problem==
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Suppose that <math>x</math> and <math>y</math> are real numbers such that <math>\log_x 3y = \tfrac{20}{13}</math> and <math>\log_{3x}y=\tfrac23</math>.  The value of <math>\log_{3x}3y</math> can be expressed in the form <math>\tfrac ab</math> where <math>a</math> and <math>b</math> are positive relatively prime integers.  Find <math>a+b</math>.
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==Corrected Solution and Answer==
 
==Corrected Solution and Answer==
  
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Solution by D. Adrian Tanner
 
Solution by D. Adrian Tanner
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(Original solution and answer below - origin unknown)
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(Original solution and answer below)
  
  
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***********************************************************************
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==Original Solution==
  
 
By rearranging the values, it is possible to attain an  
 
By rearranging the values, it is possible to attain an  
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y= 3^ (33/17)
 
y= 3^ (33/17)
  
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Therefore, a/b is    equal to 25/61,  so 25+41= 066
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Therefore, a/b is    equal to 25/61,  so 25+41= 061

Revision as of 22:52, 10 June 2015

Problem

Suppose that $x$ and $y$ are real numbers such that $\log_x 3y = \tfrac{20}{13}$ and $\log_{3x}y=\tfrac23$. The value of $\log_{3x}3y$ can be expressed in the form $\tfrac ab$ where $a$ and $b$ are positive relatively prime integers. Find $a+b$.


Corrected Solution and Answer

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)


Original Solution

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= 061

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