Question

lod2(2x+1)>log2(x-1)

Answer 1

ОДЗ :
$\left \{ {{2x+1\ \textgreater \ 0} \atop {x-1\ \textgreater \ 0}} \right. \\\\ \left \{ {{x\ \textgreater \ -0,5} \atop {x\ \textgreater \ 1}} \right. \\\\x\ \textgreater \ 1\\\\\\log _{2}(2x+1)\ \textgreater \ log _{2}(x-1)\\\\2x+1\ \textgreater \ x-1\\\\x\ \textgreater \ -2$
x ∈ (- 2 ; + ∞)
С учётом ОДЗ окончательный ответ: 
x ∈ (1 ; + ∞)