Answer
Solution set = $\emptyset $
Work Step by Step
For the equation to be defined, logarithms must have positive arguments
$\left[\begin{array}{llll}
2x-1 \gt 0, & and & x+3 \gt 0 & \\
x \gt 1/2 & & x\gt-3 &
\end{array}\right]$
$ x \gt 1/2 \qquad (*)$
This is the condition eventual solutions must satisfy.
$\log(2x-1)=\log(x+3)+\log 3\qquad $... RHS: product rule
$\log(2x-1)=\log[(x+3)\cdot 3]\quad $ ... $\log $ is one-to-one
$2x-1=3x+9$
$-1-9=3x-2x $
$-10=x\quad $ ... does not satisfy (*)
Thus, the solution set = $\emptyset $
