solution: Working on the first equation, we have from the difference of squares factorization that $\log_6 (x-y) + \log_6 (x+y) = \log_6 (x^2-y^2) = 2$, so $x^2 - y^2 = 6^2 = 36$. Using the change of base formula, the second equation becomes $$\frac{\log(5x)}{\log y} = 2 \Longrightarrow \log(5x) = 2\log y = \log y^2.$$Substituting that $y^2 = x^2 - 36$, it follows that $\log (x^2 - 36) = \log y^2 = 2\log y = \log 5x$. Since the logarithm is a one-to-one function, it follows that $x^2 - 36 = 5x$, so $x^2 - 5x - 36 = (x - 9)(x + 4) = 0$. Thus, $x = 9, - 4$, but the second does not work. Thus, our answer is $x = \boxed{9}$.
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  If $\log_6 (x-y) + \log_6 (x+y) = 2$ and $\log_y 5x = 2$, then find $x$.
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: 9
extra_info.question: If $\log_6 (x-y) + \log_6 (x+y) = 2$ and $\log_y 5x = 2$, then find $x$.
extra_info.split: test