Question 446447
ln(x − 6) + ln(x + 7) = 1 
ln(x − 6)(x + 7) = 1 
(x − 6)(x + 7) = e^1 
x^2+x-42 = e^1 
x^2+x-42-e^1 = 0 
x^2+x-42-2.718282 = 0 
x^2+x-44.718282 = 0 
apply the quadratic formula to get:
x = {6.2058, -7.2058}
the negative solution is extraneous -- throw it out leaving:
x = 6.2058
.
Details of quadratic follows:
*[invoke quadratic "x", 1, 1, -44.71828183 ]