Answer
Solution set = $\displaystyle \{\frac{11}{3}\}.$
Work Step by Step
For the equation to be defined, logarithms must have positive arguments
$\left[\begin{array}{lll}
x-2 \gt 0, & \Rightarrow & x \gt 2\\
x+3\gt 0, & \Rightarrow & x \gt -3\\
x-1 \gt 0, & \Rightarrow & x \gt 1\\
x+7 \gt 0, & \Rightarrow & x \gt -7
\end{array}\right]$
$ x \gt 2 \qquad (*)$
This is the condition eventual solutions must satisfy.
$\ln(x-2)-\ln(x+3)=\ln(x-1)-\ln(x+7)$
We use the quotient rule on both sides:
$\ln \displaystyle \frac{x-2}{x+3}=\ln\frac{x-1}{x+7}$
Recall that ln is one to one:
$\displaystyle \frac{x-2}{x+3}=\frac{x-1}{x+7}$
Multiply with the LCD, $(x+3)(x+7)$:
$(x-2)(x+7)=(x+3)(x-1)$
$ x^{2}+5x-14=x^{2}+2x-3$
$5x-2x=-3+14$
$3x=11$
$ x=\displaystyle \frac{11}{3} \qquad $ ... satisfies (*)
Thus, the solution set = $\displaystyle \{\frac{11}{3}\}.$
