Solution for $\ln{(x-1)}+\ln{(x-2)}=3$

please check solution, for this assingment. Thans.

$$\ln{(x-1)}+\ln{(x-2)}=3\\ \text{adding two ln}\\ \ln{((x-1)*(x-2))}=3\\ (x-1)*(x-2)=e³\\ x²-2x-x+2=e³\\ x²-3x+2-e³=0\\ \text{quadratic equation where: a=1, b=-3, c=2-e³}\\ \frac{-b\pm\sqrt{b²-4ac}}{2a}\\ \frac{3\pm\sqrt{9-4*1*(2-e³)}}{2*1}\\ \frac{3\pm\sqrt{9-8+4e³}}{2}\\ x=\frac{3\pm\sqrt{1+4e³}}{2}\\ \text{so now I have that this is the final equation. Is this right???}$$ Thanks???

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I don't see anything wrong... Do you have any questions? – kaine Jan 2 '14 at 18:07
Your work looks okay. Bear in mind that neither $x-1$ nor $x-2$ should be negative, so you may need the discard one of the two solutions to the quadratic equation you solved. – yoknapatawpha Jan 2 '14 at 18:07
What's wrong with $x - 1$ or $x - 2$ negative? The problem didn't say to only look at real solutions. – Ayesha Jan 2 '14 at 18:13
When using log of negative numbers or complex numbers, one can't talk of the log in a way that is consistent. – user44197 Jan 2 '14 at 18:17
You would assume that the problem is talking about $Log(z)$ in that case? But I see what you mean, thank you. – Ayesha Jan 2 '14 at 18:22

Well, not yet! By the domain of the natural logarithm, $x> 2$. So you have to verify whether your solutions are greater than $2$. If one of them is not, then you should discard it.
one of them is $\approx +6$ and the other $\approx -3$. So reject the negative solution. – user44197 Jan 2 '14 at 18:12