Question

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

Answer 1

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

Explanation:

We can change the sum of logs into a product of arguments as:
`ln[x(x-2)]=1`
We then use the definition of log to get:
`x(x-2)=e^1`
Rearranging:
`x^2-2x-e=0`
Use the Quadratic Formula:
`x_(1,2)=(2+-sqrt(4+4e))/2=(2+-sqrt(4(1+e)))/2=`
`=(2+-2sqrt(1+e))/2=(2(1+-sqrt(1+e)))/2=1+-sqrt(1+e)`
So:
`x_1=1+sqrt(1+e)`
`x_2=1-sqrt(1+e)` NO, because negative.