Log2(2x+1) = log2(x-2)

$\displaystyle log_2(2x+1)=log_2(x-2)$

ОДЗ:
$\displaystyle \left \{ {{2x+1\ \textgreater \ 0} \atop {x-2\ \textgreater \ 0}} \right. \rightarrow \left \{ {{2x\ \textgreater \ -1} \atop {x\ \textgreater \ 2}} \right. \rightarrow \left \{ {{x\ \textgreater \ -0.5} \atop {x\ \textgreater \ 2}} \right. \rightarrow x\ \textgreater \ 2$

$\displaystyle log_2(2x+1)=log_2(x-2)\\\\log_2(2x+1)-log_2(x-2)=0\\\\log_2(\frac{2x+1}{x-2})=0\\\\\frac{2x+1}{x-2}=2^0\\\\\frac{2x+1}{x-2}-1=0\\\\2x+1-(x-2)=0\,\,\,\,(x \neq 2)\\\\2x+1-x+2=0\\x+3=0\\x=-3$

Этот корень не удовлетворяет ОДЗ (x>2) => решений нет.