ln(x-5)+ln(x+4)=ln36

are we solving for x ??

if so, then
ln[(x-5)(x+4)] = ln36
"anti-ln" it

(x-5)(x+4) = 36
x^2 - x -56 = 0
(x-8)(x+7) = 0
x = 8 or x = -7

but x=-7 would make the original undefined, so
x = 8