Question

How do you solve `lnx + ln (x-2) = 1`?

Answer 1

I found: `x=1+sqrt(1+e)`

Explanation:

I would write it as one log (multiplying the arguments) as:
`ln[x(x-2)]=1`
then use the definition of log and write:
`x(x-2)=e^1`
`x^2-2x-e=0`
use the Quadratic Formula to find `x`:
`x_(1,2)=(2+-sqrt(4+4e))/2=(2+-2sqrt(1+e))/2=1+-sqrt(1+e)`
I would use the positive one only, i.e.:
`x=1+sqrt(1+e)`