solution: First, note that if $k < 0,$ then $\log(kx)$ is defined for $x \in (-\infty, 0),$ and is strictly decreasing on that interval. Since $2\log(x+2)$ is defined for $x \in (-2, \infty)$ and is strictly increasing on that interval, it follows that $\log(kx) = 2\log(x+2)$ has exactly one real solution, which must lie in the interval $(-2, 0).$ Therefore, all the values $k = -500, -499, \ldots, -2, -1$ satisfy the condition.
  
  If $k = 0,$ then the left-hand side is never defined, so we may assume now that $k > 0.$ In this case, converting to exponential form, we have \[ kx = (x+2)^2\]or \[x^2 + (4-k)x + 4 = 0.\]Any solution of this equation satisfies $\log(kx) = 2\log(x+2)$ as well, as long as the two logarithms are defined; since $k > 0,$ the logarithms are defined exactly when $x > 0.$ Therefore, this quadratic must have exactly one positive root.
  
  But by Vieta's formulas, the product of the roots of this quadratic is $4,$ which is positive, so the only way for it to have exactly one positive root is if it has $\sqrt{4} = 2$ as a double root. That is, \[x^2 + (4-k)x + 4 = (x-2)^2 = x^2 - 4x + 4\]for all $x,$ so $4-k=-4,$ and $k=8,$ which is the only positive value of $k$ satisfying the condition.
  
  In total, there are $500 + 1 = \boxed{501}$ values of $k$ satisfying the condition.
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  Find the number of integer values of $k$ in the closed interval $[-500,500]$ for which the equation $\log(kx)=2\log(x+2)$ has exactly one real solution.
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: 501
extra_info.question: Find the number of integer values of $k$ in the closed interval $[-500,500]$ for which the equation $\log(kx)=2\log(x+2)$ has exactly one real solution.
extra_info.split: test