Natural Log Help Please: Solve ln(x) + ln(x - 1) = 1 for x

[mrpersonality]

I am working on a problem and am stuck. I feel like my problem could be due to a lack of knowledge in arithmetic but I am not certain. The fact that there is an x^2 and an x is really 'throwing me off'.

The problem is:
Solve the equation for x.
ln(x) + ln(x - 1) = 1

The steps I took before I became stuck were as follows:
ln(x) + ln(x - 1) = 1
ln(x^2 - x) = 1 <==Log rule of addition
x^2 - x = e^1 <==Natural log simplifying
x(x - 1) = e^1 <==Factored

Please help. Thanks!

[stapel_eliz]

Solve the equation for x.
ln(x) + ln(x - 1) = 1

The steps I took before I became stuck were as follows:
ln(x) + ln(x - 1) = 1
ln(x^2 - x) = 1 <==Log rule of addition
x^2 - x = e^1 <==Natural log simplifying

Now work with the resulting quadratic equation (since "e" is just another number):

. . . . .

This can be solved with the Quadratic Formula.

[mrpersonality]

I didn't think to use the quadratic formula. Thank you!

[1 +- sqrt(1 - 4*1*(-e))]/2
1 + sqrt(1 + 4*e)/2
2.2229