Strange logarithm inequality

Question

I was playing around with this function: $$f(x)=\log_{\frac{1}{2}}(x^2-x-6)-\log_{\frac{1}{2}}\frac{x+2}{x-3}+4$$ I tried to find values of $x$ such that $f(x)\leq 0$. But for some reason, the answer I obtain by solving this inequality analytically does not match the answer I find by plotting this function. Here are my steps:
First I rewrite both logarithms using their properties $$\log_{\frac{1}{2}}((x+2)(x-3))-\log_{\frac{1}{2}}(x+2)+\log_{\frac{1}{2}}(x-3)+4\leq0$$ $$\log_{\frac{1}{2}}(x+2)+\log_{\frac{1}{2}}(x-3)-\log_{\frac{1}{2}}(x+2)+\log_{\frac{1}{2}}(x-3)+4\leq0$$ $$2\log_{\frac{1}{2}}(x-3)+4\leq0\implies 2\log_{\frac{1}{2}}(x-3)\leq -4$$ I rewrite the base of the logarithm $$-\log_{2}(x-3)\leq -2\implies \log_{2}(x-3)\geq 2$$ $$x-3\geq4\implies x\geq7$$ Then I find the intersection of my solution with the condition that arguments of logarithms in the initial function have to be both positive, which turns out to be $x\in(-\infty;-2)\cup(3;+\infty)$. The result of this action is $x\geq 7$.

However, when I plot $y=f(x)$, I see this solution: $x\in(-\infty;-2)\cup[7;+\infty)$.

I'm not sure I see what I am somehow missing in my solution. Why are the answers different? What is my solution missing?
Thank you in advance :)