Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Corrected Solution and Answer)
m
Line 2: Line 2:
  
 
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>.
−
 
  
 
==Corrected Solution and Answer==
 
==Corrected Solution and Answer==
Line 20: 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==
 
==Original Solution==

Revision as of 22:54, 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 $100+99=\boxed{199}.$

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_I_2015_Problems/Problem_2&oldid=70735"