13. Given $a>0, a \neq 1$, find the range of $k$ for which the equation $\log _{\sqrt{a}}(x-a k)=\log _{a}\left(x^{2}-a^{2}\right)$ has a solution.
13. $\log _{a}(x-a k)=\log _{a} \sqrt{x^{2}-a^{2}}$, then $\sqrt{x^{2}-a^{2}}=x-a k>0$, we get $k<\frac{x}{a}=\frac{1+k^{2}}{2 k}(k \neq$ 0 ). From $k<\frac{1+k^{2}}{2 k}$, we get $k<-1$ or $0<k<1$. Also, when $k=0$, from the original equation we get $a=0$, which is a contradiction.