Answer
Solution set = $\emptyset.$
Work Step by Step
For the equation to be defined, logarithms must have positive arguments
$\left[\begin{array}{llllll}
x-4 \gt 0, & and & x+1 \gt 0 & and & x-8 \gt 0 & \\
x\gt 4 & & x\gt -1 & & x\gt 8 &
\end{array}\right]$
$ x \gt 8 \qquad (*)$
This is the condition eventual solutions must satisfy.
$\ln(x-4)+\ln(x+1)=\ln(x-8)$ ... LHS: product rule
$\ln(x^{2}-3x-4)=\ln(x-8)$ ... $\log $ is one-to-one
$ x^{2}-3x-4=x-8$
$ x^{2}-4x+4=0$
We recognize a perfect square.
$(x-2)^{2}=0$
$ x=2 \quad $ ... does not satisfy (*)
Thus, the solution set = $\emptyset.$
