Answer
No solution or $\varnothing $.
Work Step by Step
The given equation is
$=\ln (x-4)+ \ln (x+1)= \ln (x-8)$
Use product rule.
$=\ln (x-4)(x+1)=\ln (x-8)$
$=(x-4)(x+1)=x-8$
Use the distributive law.
$=x^2-4x+x-4=x-8$
Add $-x+8$ to both sides.
$=x^2-4x+x-4-x+8=x-8-x+8$
Simplify.
$=x^2-4x+4=0$
Factor.
$=x^2-2x-2x+4=0$
$=x(x-2)-2(x-2)=0$
$=(x-2)(x-2)=0$
$=(x-2)^2=0$
Take the square root of both sides
$x-2 = 0$
$x=2$.
To check the solution, plug $x=2$ into the given equation.
$=\ln (2-4)+ \ln (2+1)= \ln (2-8)$
$=\ln (-2)+ \ln (2+1)= \ln (-6)$
Hence, there is no solution.
