Решите пожалуйста неравенство очень срочно надо ㏒₃(5-2x)-㏒₃(25-x)>㏒₃(x+5)-2

Решите задачу:

$log_3(5-2x)-log_3(25-x)\ \textgreater \ log_3(x+5)-2\\\\ODZ:\; \left \{ {{5-2x\ \textgreater \ 0\; ,\; 25-x\ \textgreater \ 0} \atop {x+5\ \textgreater \ 0}} \right. \; \left \{ {{x\ \textless \ 2,5\; ;\; x\ \textless \ 25} \atop {x\ \textgreater \ -5}} \right. \; \; \to \; \; -5\ \textless \ x\ \textless \ 2,5\\\\log_3(5-2x)+\underbrace{log_33^2}_{2}\ \textgreater \ log_3(x+5)+log_3(25-x)\\\\log_3\Big (9\cdot (5-2x)\Big )\ \textgreater \ log_3(x+5)(25-x)\\\\9\cdot (5-2x)\ \textgreater \ (x+5)(25-x)\\\\45-18x\ \textgreater \ 25x-x^2+125-5x\\\\x^2-38x-80\ \textgreater \ 0\\\\D/4=441=21^2\; ,\\\\ x_1= 19-21=-2\in ODZ\; ,\\\\ x_2=19+21=40\notin ODZ$

$+++(-2)---(40)+++\\\\x\in (-\infty ,-2)\cup (40,+\infty )\\\\ \left \{ {{x\in (-\infty ,-2)\cup (40,+\infty )} \atop {x\in (-5\, ;\, 2,5)}} \right. \; \; \to \; \; x\in (-5,-2)\\\\Otvet:\; \; x\in (-5,-2)\; .$