Logarithm with quadratic solution 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? 
 A: 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.
A: $$
\begin{align}
x^2 - x - e & = 1\cdot x^2 + (-1)\cdot x + (-e) \\  \\
& = ax^2 + bx + c.
\end{align}
$$
So $a=1$, $b= -1$, $c= -e$.
A: 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 $a = 1, b = -1, c = -e$.
A: 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 $-e$.
