2. If the equation $\lg (k x)=2 \lg (x+1)$ has only one real solution, then the range of values for $k$ is $\qquad$ .
2. The original equation is equivalent to
$$
\left\{\begin{array}{l}
k x>0 \\
x+1>0 \\
k x=(x+1)^{2}
\end{array}\right.
$$
$y_{1}=k x(>0)$ is the part of the line above the $x$-axis, $y_{2}=$ $(x+1)^{2}$ is a parabola with its vertex at $(-1, 0)$ and opening upwards.

Thus, the problem is converted to finding the range of $k$ such that $y_{1}$ and $y_{2}$ have only one intersection point when $x>-1$. As shown in the figure, when $y_{1}$ is tangent to $y_{2}$, $k=4$ or $k<0$, the two curves have only one intersection point. Therefore, the range of $k$ is
$$
k=4 \text { or } k<0 \text {. }
$$