Log0.7(x2-4x-5)=log0.7(7-3x)

Решите задачу:

$log_{0.7} (x^2-4x-5)=log_{0.7} (7-3x) \\ \left \{ {{x^2-4x-5\ \textgreater \ 0} \atop {7-3x\ \textgreater \ 0}} \right. ; \left \{ {{(x-5)(x+1)\ \textgreater \ 0} \atop {3x\ \textless \ 7}} \right. ;\left \{ {{x\in(-\infty;-1)\cup(5;+\infty)} \atop {x\in(-\infty;\frac{7}{3})}} \right. ; x\in(\infty;-1) \\ x^2-4x-5=7-3x \\ x^2-x-12=0 \\ (x-4)(x+3)=0 \\ x_1=4; x_2=3 \\ x=-3$