solve the equation for x:
ln x + ln (x-3) = 0
use the product property of logs ... $\displaystyle \ln{a} + \ln{b} = \ln(ab)$
$\displaystyle \ln[x(x-3)] = 0$
change to an exponential equation ...
$\displaystyle x(x-3) = e^0$
$\displaystyle x(x-3) = 1$
solve the quadratic equation ... remember that only values of x > 3 are valid.
ln(x) +l(x-3) =0
ln[x(x-3)] =0
x(x-3) = e^0
x(x-3)=1
x^2 - 3x -1=0 (Use the quadratic formula to find the roots)