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?
$\ln(x^2 - 1) = \ln((x + 1)(x - 1)) = \ln(x + 1) + \ln(x - 1), \tag 1$

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 $e &gt; 2.5 \Longrightarrow e^2 &gt; 6.25 \Longrightarrow e^2 - 1 &gt; 5.5$; also, $e &gt; 2.5 \Longrightarrow e + 1 &gt;3.5 \; \text{and} \; e - 1 &gt; 1.5$ 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 . . .
$\ln(x^2−1)−\ln(x−1)=2$

$\ln(x^2−1)=2+\ln(x−1)$

$e^{\ln(x^2−1)}=e^{2+\ln(x−1)}$

$e^{\ln(x^2−1)}=e^2e^{\ln(x−1)}$

$x^2−1=(e^2)(x−1)$

$x^2+(-e^2)x+(e^2-1)=0$

Now it's a quadratic

OR

$x^2−1=(e^2)(x−1)$

$(x+1)(x-1)=(e^2)(x−1)$

$x+1=e^2$

$x=e^2-1$