Помагите решать логарифмы

Решите задачу:

$log_3(x^2-4x-5)=log_3(7-3x)\; ,\\\\ODZ:\; \left \{ {{x^2-4x-5\ \textgreater \ 0} \atop {7-3x\ \textgreater \ 0}} \right. , \left \{ {{(x+1)(x-5)\ \textgreater \ 0} \atop {x\ \textless \ \frac{7}{3}}} \right. ,\; \left \{ {{x\in (-\infty,-1)U(5,+\infty)} \atop {x\ \textless \ \frac{7}{3}}} \right. ,\; x\ \textless \ -1\\\\x^2-4x-5=7-3x\\\\x^2-x-12=0\\\\x_1=-3,\; x_2=4\ \textgreater \ -1\; \; (teor.\; Vieta)\\\\Otvet:\; x=-3$