jmc

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 $k=-500,-499,\dots ,-2,-1$ 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.