6. If the equation $\lg k x=2 \lg (x+1)$ has only one real root, then the range of values for $k$ is $\qquad$ .
Solve:
$$
\lg k x=2 \lg (x+1) \Rightarrow k x=(x+1)^{2} \Rightarrow k=x+\frac{1}{x}+2 \text { . }
$$

When $k>0$, $x>0, k=x+\frac{1}{x}+2 \geqslant 4 \Rightarrow k=4$;
When $k<0$, $-1<x<0, k=x+\frac{1}{x}+2<0$.
Conversely, when $k=4$, the line $y=4 x$ is tangent to $y=(x+1)^{2}$, the equation has only one positive real root; when $k<0$, the equation $k x=(x+1)^{2} \Rightarrow x^{2}+(2-k) x+1=0$ must have two negative roots and $x_{1}<x_{2}$, thus by $x_{1} x_{2}=1 \Rightarrow x_{1} \in(-\infty,-1), x_{2} \in(-1,0)$ satisfying the condition.

Therefore, the range of $k$ is $(-\infty, 0) \bigcup\{4\}$.