Answer :
[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]