Могите решить log2(3x+7)>log2(5x-1)

$\left \{ {{3x+7\ \textgreater \ 0} \atop {5x-1\ \textgreater \ 0}} \right. \\ \left \{ {{3x\ \textgreater \ -7} \atop {5x\ \textgreater \ 1}} \right. \\ \left \{ {{x\ \textgreater \ -7/3} \atop {x\ \textgreater \ 1/5}} \right. \\ 3x+7\ \textgreater \ 5x-1 \\ 3x-5x\ \textgreater \ -7-1 \\ -2x\ \textgreater \ -8 \\ x\ \textless \ 4 \\ x \ prinadlegit\ (1/5;4)$