Answer
Solution set = $\emptyset $
Work Step by Step
For the equation to be defined, logarithms must have positive arguments
$\left[\begin{array}{llll}
3x-3 \gt 0, & and & x+1 \gt 0 & \\
x \gt 1 & & x\gt-1 &
\end{array}\right]$
$ x \gt 1 \qquad (*)$
This is the condition eventual solutions must satisfy.
$\log(3x-3)=\log(x+1)+\log 4\qquad $... RHS: product rule
$\log(3x-3)=\log[(x+1)\cdot 4]\quad $ ... $\log $ is one-to-one
$3x-3=4x+4$
$-3-4=4x-3x $
$-7=x\quad $ ... does not satisfy (*)
Thus, the solution set = $\emptyset $
