$$\bbox[aqua]{\log _{\frac{x}{2}}(x+2)-\log _x(4-x)=1}$$
$$\mapsto \log _{\frac{x}{2}}(x+2)=\log _x x+\log _x(4-x) \\\Rightarrow \log _{\frac{x}{2}}(x+2)=\log _x(x(4-x)) \\\Rightarrow \frac{\log _x(x+2)}{\log _x\left(\frac{x}{2}\right)}=\log _x(x(4-x)) \\\Rightarrow \log _x(x+2)=\log _x(x(4-x)) \log _x\left(\frac{x}{2}\right) \\\Rightarrow \color{\red}{x+2=x^{\log _x(x(4-x)) \log _x\left(\frac{x}{2}\right)}} \Rightarrow \, ? ? ? $$
The equation is marked in blue, I also wrote my solution attached to the equation, but when I got to the red part, I ran into trouble from there on and could not find a way to find the answer to $x$. I would like to know what your solution is.