Решите пожалуйста! 2 log 0,5x=log0,5(2x^2-x)

Решите задачу:

$2log_{0,5}x=log_{0,5}(2x^2-x)\; ,\; \; ODZ:\; \left \{ {{x\ \textgreater \ 0} \atop {2x^2-x\ \textgreater \ 0}} \right. \; \left \{ {{x\ \textgreater \ 0} \atop {x(2x-1)\ \textgreater \ 0}} \right. \; \\\\ODZ:\; \left \{ {{x\ \textgreater \ 0} \atop {x\in (-\infty ,0)\cup (\frac{1}{2},+\infty )}} \right. \; \Rightarrow \; \; x\in (\frac{1}{2},+\infty )\\\\log_{0,5}x^2=log_{0,5}(2x^2-x)\\\\x^2=2x^2-x\\\\x^2-x=0\\\\x(x-1)=0\\\\x_1=0\notin ODZ\; \; ;\; \; x_2=1\in ODZ\\\\Otvet:\; x=1.$