Respuesta :

konrad509
[tex]D:5x\ \textgreater \ 0 \wedge 2x+9\ \textgreater \ 0\\ D:x\ \textgreater \ 0 \wedge 2x\ \textgreater \ -9\\ D:x\ \textgreater \ 0 \wedge x\ \textgreater \ -\dfrac{9}{2}\\ D:x\ \textgreater \ 0\\\\ \log5x=\log(2x+9)\\ 5x=2x+9\\ 3x=9\\ x=3[/tex]