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. – André Nicolas Apr 30 '12 at 3:58
Beautiful, thanks. – Kurt Apr 30 '12 at 4:01
Guys, why would we need so many duplicates in such a simple question which such a direct answer? – Pedro Tamaroff Apr 30 '12 at 4:03
@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. – Alex Becker Apr 30 '12 at 4:40
@AlexBecker You're right. However, now we have the advantage that if a question is answered during a composition, the composer will be notified. – Pedro Tamaroff Apr 30 '12 at 4:43

\begin{align} x^2 - x - e & = 1\cdot x^2 + (-1)\cdot x + (-e) \\ \\ & = ax^2 + bx + c. \end{align}

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

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The quadratic formula is used to solve equations in the form $ax^2 + bx + c = 0$. In order to use the formula, you need to plug in $a, b, c$ which are coefficients not variables.

So in your case: $0 = x^2 - x - e$ can we rewritten as $0 = (1)x^2 + (-1)x + -e$. Comparing that to $0 = ax^2 + bx + c$ we see that $a = 1, b = -1, c = -e$.

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Assuming that the unknown is $x$, the $a$ and $b$ of the quadratic formula are the coefficients of $x^2$ and $x$, respectively, and $c$ is the constant term; they do not include the unknown. You have the quadratic equation $x^2-x-e=0$; this can be written $1\cdot x^2+(-1)x+(-e)=0$, so the coefficient of $x^2$ is $1$, the coefficient of $x$ is $-1$, and the constant term is $-e$.

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If one has a polynomial equation of the form

$$ax^2+bx+c=0$$

the solutions are given by

$$x=\frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$$

Your problem is you're extracting $x^2$ as the coefficient, when the coefficient is really $1$, and the same is going on with $x$. Note that

$$x^2+x-e=0$$ is the same as

$$\color{red}{1}x^2+\color{blue}{1}x+\color{green}{(-e)}=0$$

So what you have to plug in the equation is

$$x=\frac{{ - \color{blue}{1} \pm \sqrt {{\color{blue}{1}^2} - 4\cdot \color{red}{1}\cdot \color{green}{(-e)}} }}{{2\cdot \color{red}{1}}}$$

which will give you the solutions you are looking for.

Note that we must also have that $x>1$, since $\ln (x-1)$ is not defined for $x\leq 1$. As a consequence, if any value of $x$ is smaller than $1$, we can't take it as a solution.

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