9. Given $a>0, a \neq 1$, find the range of $k$ for which the equation $\log _{a}(x-a k)=\log _{a^{\prime}}\left(x^{2}-a^{2}\right)$ has a solution.
9. The original equation is equivalent to
$$
\left\{\begin{array}{l}
x-a k>0, \\
x^{2}-a^{2}>0, \\
(x-a k)^{2}=x^{2}-a^{2} .
\end{array}\right.
$$

has a solution. It is also equivalent to
$$
\left\{\begin{array}{l}
x-a k>0 \\
(x-a k)^{2}=x^{2}-a^{2}
\end{array}\right.
$$

From (5), $2 a k x=a^{2}\left(1+k^{2}\right), x=\frac{a^{2}\left(1+k^{2}\right)}{2 a k}$, substituting into (4), it is easy to get $k \in(0,1) \cup(-\infty,-1)$.