Answer
Solution set = $\emptyset.$
Work Step by Step
For the equation to be defined, logarithms must have positive arguments
$\left[\begin{array}{lll}
x-5 \gt 0, & \Rightarrow & x \gt 5\\
x+4\gt 0, & \Rightarrow & x \gt -4\\
x-1 \gt 0, & \Rightarrow & x \gt 1\\
x+2 \gt 0, & \Rightarrow & x \gt -2
\end{array}\right]$
$ x \gt 5 \qquad (*)$
This is the condition eventual solutions must satisfy.
$\ln(x-5)-\ln(x+4)=\ln(x-1)-\ln(x+2)$
Apply the quotient rule on both sides:
$\displaystyle \ln\frac{x-5}{x+4}=\ln\frac{x-1}{x+2}$
Recall that ln is one to one:
$\displaystyle \frac{x-5}{x+4}=\frac{x-1}{x+2}$
Multiply with the LCD, $(x+4)(x+2)$:
$(x-5)(x+2)=(x+4)(x-1)$
$ x^{2}-3x-10=x^{2}+3x-4$
$-3x-3x=-4+10$
$-6x=6$
$ x=-1 \quad $ ... does not satisfy (*)
Thus, the solution set = $\emptyset.$
