13. For what value of $a$ does the equation $\log _{x+a} 2 x=2$ have two distinct real roots?
13. The original equation is equivalent to $\left\{\begin{array}{l}(x+a)^{2}=2 x, \\ x>0, \\ x+a>0 \\ x+a \neq 1\end{array} \Leftrightarrow\left\{\begin{array}{l}x^{2}+2(a-1) x+a^{2}=0, \\ x \neq 0, \\ x>-a \\ x \neq 1-a\end{array}\right.\right.$

Obviously, $\Delta=4(a-1)^{2}-4 a^{2}>0, a-a \\ 1-a-\sqrt{1-2 a} \neq 1-a\end{array}\right.$ Solving this gives $a>0$, thus $a \in\left(0, \frac{1}{2}\right)$.