Question

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

Answer 1

I found (depending upon the base `b` of your logs):
`x=(b+1)/(b-1)`

Explanation:

We do not know the base of the logs (it could be `10`...) so we say that the base is `b`:
we get:
`log_b(x+1)-log_b(x-1)=1`
we use the property of logs that tells us:
`logx-logy=log(x/y)`
and write:
`log_b((x+1)/(x-1))=1`
we now use the definition of log:
`log_bx=a ->x=b^a`
`(x+1)/(x-1)=b^1`
rearranging:
`x+1=b(x-1)`
`x+1=bx-b`
`x-bx=-1-b`
`x(1-b)=-1-b`
`x=(-1-b)/(1-b)=-(b+1)/(1-b)=(b+1)/(b-1)`

Now you can choose the base `b` of your original logs and find `x`.
If, for example, `b` were `10` then you get:
`x=11/9`