Question

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

Answer 1

I found: `x=(2+sqrt(4+4e))/2=1+sqrt(1+e)`

Explanation:

We can change the sum of logs into a log of a product and write:
`ln[x(x-2)]=1`
use the definition of log:
`x(x-2)=e^1`
rearrange and solve for `x`:
`x^2-2x-e=0`
Use the Quadratic Formula:
`x_(1,2)=(2+-sqrt(4+4e))/2`
`x_1=(2+sqrt(4+4e))/2>0` YES
`x_2=(2-sqrt(4+4e))/2<0` NO