[tex]\bf \textit{logarithm of factors}\\\\
log_{{ a}}(xy)\implies log_{{ a}}(x)+log_{{ a}}(y)\\\\
-------------------------------\\\\
ln(x-6)+ln(x+10)=ln(x-15)
\\\\\\
ln[(x-6)(x+10)]=ln(x-15)\implies (x-6)(x+10)=x-15
\\\\\\
x^2-5x-6=x-15\implies x^2-6x+9=0
\\\\\\
(x-3)^2=0\implies \boxed{x=3}[/tex]
ok.. now, if we use that on say ln(x-15), we end up with [tex]\bf ln(-12)\implies log_e(-12)[/tex].
now, there's no exponent whatsoever, that would give us a negative result for a base of "e", namely, x = 3 is NOT in the domain for ln(), so the system is inconsistent, or has no solution.
