Question 28, Given $k \in[-500,500]$, and $k \in Z$, if the equation $\lg (k x)=2 \lg (x+2)$ has exactly one real root, then the number of $k$ that satisfies the condition is $\qquad$ —.
Question 28, Answer: If $\mathrm{k}0$, and $\mathrm{x}+2>0$, we know $\mathrm{x} \in(-2,0)$. At this time, the equation can be transformed into $k=x+\frac{4}{x}+4$. Note that $f(x)=x+\frac{4}{x}+4$ is monotonically decreasing in $(-2,0)$, and its range is $(-\infty, 0)$. Therefore, each integer in the interval $[-500,0)$ satisfies the condition.

If $k>0$, then by $k x>0$, and $x+2>0$, we know $x \in(0,+\infty)$. At this time, the equation can be transformed into $k=x+$ $\frac{4}{x}+4$. Note that $f(x)=x+\frac{4}{x}+4$ reaches its minimum value 8 at $x=2$. When $k>8$, the equation $k=x+\frac{4}{x}+4$ has two distinct real roots. Therefore, when $k>0$, the only integer that satisfies the condition is $k=8$.
In summary, the number of $\mathrm{k}$ that satisfy the condition is 501.