Question

How do you solve `ln(x+1) - 1 = ln(x-1)`?

Answer 1

I found: `x=-(1+e)/(1-e)`

Explanation:

I would rearrange your equation as:
`ln(x+1)-ln(x-1)=1`
now I use the fact that:
`loga-logb=log(a/b)`
to get:
`ln((x+1)/(x-1))=1`
we can take the exponential of both sides:
`e^(ln((x+1)/(x-1)))=e^1` eliminating the `ln`;
`(x+1)/(x-1)=e`
`x+1=ex-e`
rearranging:
`x-ex=-1-e`
`x(1-e)=-(1+e)`
`x=-(1+e)/(1-e)`