Question 704682
ln(x)+ln(x-8)=5 
ln((x)(x-8)) = 5
(x)(x-8) = e^5
x^2-8 = e^5
x^2-8x = e^5
x^2-8x-e^5 = 0
Solving using the "quadratic formula" yields:
x = {16.82, -8.82}
Since you can't take the ln of a negative number, the -8.82 is an extraneous solution.  Throw it out leaving:
x = 16.82
.
Details of quadratic follows:
*[invoke quadratic "x", 1, -8, -e^5 ]