Question

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

Answer 1

Do some algebra and apply some log properties to get `x=-(e+1)/(1-e)~~2.164`

Explanation:

Begin by collecting the `x` terms on one side of the equation and the constant terms (numbers) on the other:
`ln(x+1)-ln(x-1)=1`

Apply the natural log property `lna-lnb=ln(a/b)`:
`ln((x+1)/(x-1))=1`

Raise both sides to the power of `e`:
`e^ln((x+1)/(x-1))=e^1`
`->(x+1)/(x-1)=e`

This equation is solved in a few steps:
`x+1=e(x-1)`
`x+1=ex-e`
`x-ex=-e-1`
`x(1-e)=-e-1`
`x=-(e+1)/(1-e)~~2.164`

Note that `(-e-1)/(1-e)=(-1(e+1))/(1-e)=-(e+1)/(1-e)`.