6 If the equation $\lg k x=2 \lg (x+1)$ has only one real root, then the range of values for $k$ is $\qquad$ .
$$
6 k0, \\ x+1>0, \\ k x=(x+1)^{2},\end{array}\right.
$$
if and only if
$$
\begin{array}{c}
k x>0, \\
x+1>0, \\
x^{2}+(2-k) x+1=0 .
\end{array}
$$

For (3), by the quadratic formula we get
$$
\begin{array}{l}
x_{1}, x_{2}=\frac{1}{2}\left[k-2 \pm \sqrt{k^{2}-4 k}\right], \\
\Delta=k^{2}-4 k \geqslant 0 \Rightarrow k \leqslant 0 \text { or } k \geqslant 4 .
\end{array}
$$
(i) When $k0,
\end{array}\right.
$$

Thus, $x_{1}$ and $x_{2}$ are both negative roots.
From (4) we know
$$
\left\{\begin{array}{l}
x_{1}+1>0, \\
x_{2}+14$ we get from (3)
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=k-2>0, \\
x_{1} x_{2}=1>0,
\end{array}\right.
$$

Thus, $x_{1}$ and $x_{2}$ are both positive roots, and $x_{1} \neq x_{2}$, which does not meet the requirement, so it is discarded.
In summary, $k<0$ or $k=4$ are the solutions.