Solve for x: ln x + ln(x-1) = 1?

ln x + ln(x-1) = 1
ln( x^2 - x) = 1
e = x^2 - x
x^2 - x - e = 0

Use quadratic formula
x=-1.222 and x=2.222
but cannot take log of negative number so eliminate x=-1.2222

ln(x) + ln(x-1) = 1
ln(x*(x-1)) = 1
x^2 - x = e
x^2 - x - e = 0
x = [ 1 +/- sqrt(1 + 4e) ] / 2

Since log isn't defined for negative numbers, the true solution is:

x = [ 1 + sqrt(1 + 4e) ] / 2

ln [ x ( x - 1 ) ] = 1

x ( x - 1 ) = e

x ² - x - e = 0

x ² - x - 2.78 = 0

x = [ 1 ± √ ( 1 + 11.12 ) ] / 2

x = [ 1 ± √ ( 12.12 ) ] / 2

x = [ 1 ± 3.48 ] / 2

x = 2.24 , x = - 1.24

Remember that log(a) + log(b) = log(ab) for logs of any base. So here we have:
ln(x) + ln(x-1) = 1
ln( x(x-1) ) = 1

This means x(x-1) = e^1, which you can write as
x^2 - x - e = 0

Using the quadratic formula, you get:
x = [ 1 ± √(1 - 4(1)(-e)) ] / 2
x = [ 1 ± √(1 + 4e) ] / 2

One of these answers however is a negative number (4e+1 is obviously greater than 1, so its square root must be greater than 1 too, so 1 - √(1+4e) is negative). This wouldn't make sense for the ln(x) term in the original equation because you can't take the log of a negative number. So the only solution is
x = [ 1 + √(1 + 4e) ] / 2

WHY some people insist on writing out the answer as a crappy decimal approximation, instead of just leaving the true answer as is, I have no idea. Is that what they're teaching in schools these days?

ln(x) + ln(x - 1) = 1
ln[x(x - 1)] = 1
x(x - 1) = e^1
x^2 - x = e
x^2 - x/2 - x/2 = e
x^2 - x/2 - x/2 + 1/4 = e + 1/4
(x^2 - x/2) - (x/2 - 1/4) = 4e/4 + 1/4
x(x - 1/2) - 1/2(x - 1/2) = (4e + 1)/4
(x - 1/2)^2 = (4e + 1)/4
x - 1/2 = ±√[(4e + 1)/4]
x = 1/2 ±[√(4e + 1)]/2
x = [1 ±√(4e + 1)]/2

The addition of logs is a multiplication inside the logs.
ln(x) + ln(x-1) = 1
ln[x*(x-1)] = 1
ln(x^2 - x) = 1
Use each as a exponent on e. This will cancel the ln on the left side.
x^2 - x = e^1
x^2 - x - e = 0
This has to go through the quadratic solution.
x = [-b +- sqrt(b^2 - 4ac)] / (2a)
x = [1 +- sqrt(1^2 - 4(1)(-e)] / (2*1)
x = [1 +- sqrt(1 + 4e)] / 2
x = [1 + sqrt(1 + 4e)] / 2 OR x = [1 - sqrt(1 + 4e)] / 2
x ~= 2.223 OR x = -1.223
~= means approximately equal to. I had to round.

Now we need to check that both values are indeed valid. Looking back at the original equation, we have ln(x) and ln(x-1). The parameter for a log must be positive (not even 0). Therefore, the negative value for x is invalid. The only acceptable answer is x ~= 2.223.

ln{[x(x-1)] = lne
x(x-1)=e
x2-x-e=0

ln x + ln(x-1) = 1
=> ln( x * (x-1) ) = 1
= > x(x-1) = e^1
=.> x^2 -x - e =0
= > x^2 - x - 2.718 =0
x= [ 1+- rt( 1 + 10.872)]/2
= (1 + - 3.446 }/2
= 2.223, - 1.223
As the log of a negative number is not defined we take only the positive value, so x= 2.233

since ln a + ln b = ln(ab)
ln x + ln(x-1) = ln(x(x-1))
ln(x^2 - x) =1
x^2 - x = e

Easy - just type "solve ln x + ln(x-1) = 1" into Wolfram Alpha (see link below).