Mock AIME I 2015 Problems/Problem 2


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

$\log_x 3y = \frac{log_3 3y}{log_3 x}  = \frac{1+log_3 y}{log_3 x} = \frac{20}{13}$


$\log_{3x} y = \frac{log_3 y}{log_3 3x}  = \frac{log_3 y}{1+log_3 x} = \frac{2}{3}.$

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

$x= 3^ {65/17}$


$y= 3^ {33/17}$

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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