Difference between revisions of "Mock AIME I 2015 Problems/Problem 2"
(Incorrect solution and answer resolved) |
Blunderbro (talk | contribs) (→Original Solution) |
||
(3 intermediate revisions by 2 users not shown) | |||
Line 1: | Line 1: | ||
+ | ==Problem== | ||
+ | |||
+ | 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>. | ||
+ | |||
==Corrected Solution and Answer== | ==Corrected Solution and Answer== | ||
Line 15: | Line 19: | ||
<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> |
− | |||
Solution by D. Adrian Tanner | Solution by D. Adrian Tanner | ||
− | (Original solution and answer below | + | (Original solution and answer below) |
− | + | ==Original Solution== | |
− | |||
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 | ||
− | + | 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 02:24, 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.