LG(5x-7)>2; log0,1(5-3x)\<-2 решение простейших логарифмических неравенств

Решите задачу:

$lg(5x-7)\ \textgreater \ 2 \\ \\lg(5x-7)\ \textgreater \ 2\cdot 1 \\ \\ 1=lg10 \\ \\lg(5x-7)\ \textgreater \ 2\cdot lg10 \\ \\lg(5x-7)\ \textgreater \ 2\cdot lg10 \\ \\ lg(5x-7)\ \textgreater \ lg10^2 \\ \\ \left \{ {{5x-7\ \textgreater \ 100} \atop {5x-7\ \textgreater \ 0}} \right. \\ \\ 5x\ \textgreater \ 107 \\ \\ x\ \textgreater \ 21,4$

$log0,1(5-3x)\ \textless \ -2 \\ \\ log0,1(5-3x)\ \textless \ -2\cdot log{0,1}0,1 \\ \\ log0,1(5-3x)\ \textless \ log{0,1}(0,1)^{-2} \\ \\ \left \{ {{5-3x\ \textgreater \ 0} \atop {5-3x\ \textgreater \ 100}} \right. \\ \\ 5-3x\ \textgreater \ 100 \\ \\ -3x\ \textgreater \ 95 \\ \\ x\ \textless \ -31 \frac{2}{3}$