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

(Created page with "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")
 
(→‎Original Solution)
 
(4 intermediate revisions by 2 users not shown)
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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==
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Use the logarithmic identity  <math>\log_q p = \frac{log_r p}{log_r q}</math> to expand the assumptions to
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<math>\log_x 3y = \frac{log_3 3y}{log_3 x}  = \frac{1+log_3 y}{log_3 x} = \frac{20}{13}</math> 
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and
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<math>\log_{3x} y = \frac{log_3 y}{log_3 3x}  = \frac{log_3 y}{1+log_3 x} = \frac{2}{3}.</math>
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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>
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The sought ratio is
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<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>
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The answer then is <math>100+99=\boxed{199}.</math>
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Solution by D. Adrian Tanner
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(Original solution and answer below)
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==Original Solution==
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By rearranging the values, it is possible to attain an  
 
By rearranging the values, it is possible to attain an  
  
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x= 3^ (65/17)
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<math>x= 3^ {65/17}</math>
  
 
and
 
and
  
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y= 3^ (33/17)
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<math>y= 3^ {33/17}</math>
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Therefore, a/b is    equal to 25/61,  so 25+41= 061
  
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Therefore, a/b is     equal to 25/61, so 25+41= 066
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Ignore the original solution as it is incorrect - blunderbro
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Solution at the top checked by blunderbro
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Note: It is correct, original one was the first one posted but was incorrect.

Latest revision as of 02:24, 9 January 2017

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 $100+99=\boxed{199}.$

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

Ignore the original solution as it is incorrect - blunderbro Solution at the top checked by blunderbro Note: It is correct, original one was the first one posted but was incorrect.

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