Log3x+logx3 больше 0 решите пожалуйста

$OD3: \begin{cases}x\ \textgreater \ 0\\x\neq1\end{cases}=\ \textgreater \ x\in(0;1)\cup(1;+\infty)$
$log_3x+log_x3\ \textgreater \ 0\\log_3x+log_x3\ \textgreater \ 0\\\frac{1}{log_x3}+log_x3\ \textgreater \ 0\\log_x^23+1\ \textgreater \ 0\\\begin{cases}\begin{cases}x\ \textgreater \ 1\\log_x^23\ \textgreater \ -1\end{cases}\\\begin{cases}0\ \textless \ x\ \textless \ 1\\log_x^23\ \textless \ -1\end{cases}\end{cases}\\1)log^2_x3=-1\\x\in\varnothing\\2)log^2_x3=-1\\x\in\varnothing$

      (1)/////////////////////>x
///////////////////////////////////////////////>x

      (0)///(1)    >x
   >x

OD3: (0)///(1)/////////////////////>x

$x\in(1;+\infty)$