Question

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

Answer 1

`x = 1/2+1/2sqrt(1+4e^2)`

Explanation:

We have:

`lnx+ln(x-1) = 2`

`:. ln(x(x-1)) = 2`

`:. x(x-1) = e^2`

`:. x^2-x- e^2 = 0`

`:. (x-1/2)^2-(1/2)^2- e^2 = 0`

`:. (x-1/2)^2 = 1/4+e^2 = 0`

`:. x-1/2 = +-sqrt(1/4(1+4e^2))`

`:. x-1/2 = +-1/2sqrt(1+4e^2)`

`:. x = 1/2+-1/2sqrt(1+4e^2)`

Strictly speaking we require that both:

`x gt 0` and `x+1 gt 0`

So that the logarithms, in the original equation exist. Consequently we can negate one solution and we have:

`x = 1/2+1/2sqrt(1+4e^2)`