Question 427145
Applying log rules:
ln x + ln (x-2) = 1
ln x(x-2) = 1
x(x-2) = e^1
x^2-2x = e^1
x^2-2x - e^1 = 0
x^2-2x - 2.718 = 0
applying the quadratic formula we get:
x = {2.928, -0.928}
you can throw out the negative solution leaving:
x = 2.928
.
details of quadratic follows:
*[invoke quadratic "x", 1, -2, -2.718 ]