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I have looked over this question several times, and I only understand the solution up to a point.

Solve the equation for $x$:

$$\ln x+\ln(x-1)=1 $$

First thing I do is apply the additive rule of logs

$$\ln(x(x-1))=1$$ $$\ln(x^2-x)=1$$ $$e^1=x^2-x$$ then setting up for quadratic $$ 0=x^2-x-e$$

Now here is where I get lost: defining the values of $a, b$, and $c$ for the quadratic. $$\begin{align*}a&=x^2\\ b&=-x\\ c&=-e \end{align*}$$

But the solution shows:

$$a=1, b=-1, c=-e$$

I am not sure the reasoning behind using these values for the quadratic?

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    The equation $x^2-x-e=0$ is an ordinary quadratic equation. There is nothing special about $e$, it is just a number, a bit bigger than $2.7$. The quadratic does not factor pleasantly, so you will use the Quadratic Formula to find the roots. They are $x=\frac{1\pm \sqrt{1+4e}}{2}$. However, one of them is no good, because $\ln$ will not be defined.2012-04-30
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    Beautiful, thanks.2012-04-30
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    Guys, why would we need so many duplicates in such a simple question which such a direct answer?2012-04-30
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    @PeterTamaroff This is a common result of multiple answers being composed simultaneously. I know I've done it, and I can think of more than a few times you've done this.2012-04-30
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    @AlexBecker You're right. However, now we have the advantage that if a question is answered during a composition, the composer will be notified.2012-04-30
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    @Peter, the place to discuss this is the meta site.2012-04-30

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