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

From equation (3) we get
$\Delta=k^{2}-4 k \geqslant 0 \Rightarrow k \leqslant 0$ or $k \geqslant 4$.
Let the two roots of equation (3) be $x_{1}$ and $x_{2}\left(x_{1} \leqslant x_{2}\right)$.
(1) When $k<0$.
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=k-2<0, \\
x_{1} x_{2}=1>0 .
\end{array}\right.
$$

Therefore, $x_{1}$ and $x_{2}$ are both negative roots, and $x_{1}<0 .
\end{array}\right.
$$

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

Therefore, $x_{1}$ 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 $k=4$ is the solution.