[tex]\displaystyle\\\log_3(x-3) \cdot \log_23+\log_2(x+2)=1+\log_5(x-1)\cdot\log_2(5)\\\\x\in (3;\infty) \\\\\frac{\log_2(x-3)}{\log_23}\cdot\log_23+\log_2(x+2)=\log_22+\frac{\log_2(x-1)}{\log_25}\cdot\log_25\\\\\log_2(x-3)+\log_2(x+2)=\log_22+\log_2(x-1)\\\\ \log_2((x-3)(x+2))=\log_2(2(x-1))\\\\\log_2(x^2-x-6)=\log_2(2x-2)\\\\x^2-x-6=2x-2\\\\x^2-3x-4=0\\\\x_1=4 \ \ \in (3;\infty)\\x_2=-1\notin (3;\infty)\\\\\boxed{x=4}[/tex]