Follow the basics

$\begin{aligned} \text{For} \quad \log_{x^2-6x+8} \left( \color{#3D99F6}{\log_{2x^2-2x-8}\left(x^2+5x \right)} \right) & = 0 \\ \Rightarrow \color{#3D99F6}{\log_{2x^2-2x-8}\left(x^2+5x \right)} & = 1 \\ \Rightarrow 2x^2-2x-8 & = x^2+5x \\ x^2 - 7x -8 & = 0 \\ (x+1)(x-8) & = 0 \end{aligned}$

$\begin{aligned} \Rightarrow x & = \begin{cases} \color{#D61F06}{-1} & \text{but }2x^2-2x-8 = -4 \text{ and } \log_{-4} \text{ is not real; } \color{#D61F06} {\text{rejected.}} \\ \color{#3D99F6}{8} & \text{and }2x^2-2x-8 = 104 \text{ and } x^2-6x+8=24; \color{#3D99F6} {\text{ accepted.}} \end{cases} \end{aligned}$

Therefore, the answer is $\boxed{8}$ .