Difference between revisions of "Mock AIME I 2015 Problems/Problem 2"
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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>. | 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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<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> | <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> | + | The answer then is <math>100+99=\boxed{199}.</math> |
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Solution by D. Adrian Tanner | Solution by D. Adrian Tanner | ||
(Original solution and answer below) | (Original solution and answer below) | ||
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==Original Solution== | ==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 | ||
| − | x= 3^ | + | <math>x= 3^ {65/17}</math> |
and | and | ||
| − | y= 3^ | + | <math>y= 3^ {33/17}</math> |
Therefore, a/b is equal to 25/61, so 25+41= 061 | Therefore, a/b is equal to 25/61, so 25+41= 061 | ||
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| + | 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. | ||
Latest revision as of 03:24, 9 January 2017
Problem
Suppose that 
and 
are real numbers such that 
and 
. The value of 
can be expressed in the form 
where 
and 
are positive relatively prime integers. Find 
.
Corrected Solution and Answer
Use the logarithmic identity 
to expand the assumptions to
and
Solve for the values of 
and 
which are respectively 
and 
The sought ratio is
The answer then is 
Solution by D. Adrian Tanner (Original solution and answer below)
Original Solution
By rearranging the values, it is possible to attain an
and
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.
