The number of non integer solutions of the equation ${\log }_{x^2-6x+8}\left({\log }_{{2x}^2-2x-8}\left(x^2+5x\right)\right)=0$ is/are

This equation is equivalent to the system 
$\{\begin{array}{l}{x}^2-6x+8>0\\ x^2-6x+8\ne 1\\ {2x}^2-2x-8>0\\ {2x}^2-2x-8\ne 1\\ {\log }_{2x^2-2x-8}(x^2+5x)>0\\ {\log }_{2x^2-2x-8}(x^2-5x)=(x^2-6x+8{)}^0=1\Rightarrow x^2+5x={2x}^2-2x-8\end{array}$
Solve the equation of the system  
$\Rightarrow \{\begin{array}{l}x<2andx>4\\ x\ne 3\pm \sqrt{2}\\ x<\frac{1-\sqrt{17}}{2}andx>\frac{1+\sqrt{17}}{2}\\ x\ne \frac{1\pm \sqrt{19}}{2}\\ x=-1,8\end{array}$
$x=-1$, does not satisfy the third relation of this system. Hence $x=8$ is only root of the original equation.