How do I solve $\ln(x^2 − 1) − \ln(x − 1) = 2$ I need to solve the following: 

$$\ln(x^2 − 1) − \ln(x − 1) = 2$$

But I am not entirely sure how to solve this. 
I know that you can combine them into a fraction but then how would you simplify from there?
 A: $\ln(x^2 - 1) = \ln((x + 1)(x - 1)) = \ln(x + 1) + \ln(x - 1), \tag 1$
assuming all the $\ln$ terms are well-defined; thus,
$\ln(x^2 - 1) - \ln(x - 1) = \ln(x + 1); \tag 2$
then
$\ln(x + 1) = 2, \tag 3$
or
$x + 1 = e^2, \tag 4$
yielding
$x = e^2 -1. \tag 5$
It is easy to see all the $\ln$ terms make sense here, since $e > 2.5 \Longrightarrow e^2 > 6.25 \Longrightarrow e^2 - 1 > 5.5$; also, $e > 2.5 \Longrightarrow e + 1 >3.5 \; \text{and} \; e - 1 > 1.5$ so all the logarithms are defined and positive.
A check:  with $x$ as in (5)
$x^2 = e^4 - 2e^2 + 1, \tag 6$
so
$x^2 - 1 = e^4 - 2e^2 =e^2(e^2 - 2) = (x + 1)(x - 1), \tag 6$
and $\ln$ takes over from here . . . we get
$\ln(x^2 - 1) = \ln(e^2(e^2 - 2)) = \ln e^2 + \ln(e^2 - 2) = 2 \ln e + \ln(x - 1), \tag 7$
etc . . . 
A: Step 1: Condense the logarithm: $\ln\left(\frac{x^2-1}{x-1}\right)=2$ and that becomes $\frac{x^2-1}{x-1}=e^2$.(Why?) 
Now as far as the left-hand side simplification is concerned, can you let your thoughts go over that? It should become a linear equation for you to solve. 
Give it a try
A: Hint: $\ln(x)-\ln(y)=\ln(\frac{x}{y})$. Then use the fact that $e^{\ln(x)}=x$. This will eliminate all of the logs in your equation, and turn it into a manageable function of $x$.
A: $\ln(x^2−1)−\ln(x−1)=2$
$\ln(x^2−1)=2+\ln(x−1)$
$e^{\ln(x^2−1)}=e^{2+\ln(x−1)}$
$e^{\ln(x^2−1)}=e^2e^{\ln(x−1)}$
$x^2−1=(e^2)(x−1)$
$x^2+(-e^2)x+(e^2-1)=0$
Now it's a quadratic
OR
$x^2−1=(e^2)(x−1)$
$(x+1)(x-1)=(e^2)(x−1)$
$x+1=e^2$
$x=e^2-1$
