Question: > 
> $\ln(x + 1) = 2 + \ln(x - 1)$; solve for $x$.
> 
> 
> 

From there I get 
$$\ln \frac{x+1}{x-1} = 2.$$

Am I headed in the right direction, in our examples we would exponentiate both sides, does that still stand for this even though there's only a 2 on the right hand side?

Answer: 
For Q1. How about this. $X = \{0,1\}^A$ for uncountable $A$, and $\Sigma$ is the product $\sigma$-algebra. So each element of $\Sigma$ depends on only countably many coordinates. 

Now we just need a proof that this cannot be the Borel algebra of any topology.
EDIT: It's best just to ignore this nonsensical post.

The product space mentioned in the answer by Gerald Edgar does not provide a counterexample. The product $\sigma$-algebra is determined by countably many coordinate, the product topology by finitely many coordinates. A set determined by countably many coordiantes is a countable intersection of sets determined by finitely many coordinates.

Remark: This answer contained a flawed "proof" that the answer to the first question is yes. I've edited it.