Question

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

Answer 1

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

Explanation:

Before beginning, note that we do not accept values less than `0` for the `ln` function, and thus as we are beginning with `ln(x)` in the equation, it must be that `x > 0`.

We will use the following properties of logarithms:
(1) `ln(ab) = ln(a) + ln(b)`
(2) `e^(ln(a)) = a`

as well as the quadratic formula:
`ax^2 +bx + c = 0 => x = (-b+-sqrt(b^2-4ac))/(2a)`

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By (1)
`ln(x) + ln(x+1) = ln(x(x+1))=ln(x^2 + x)`
`=> ln(x^2 + x) = 2`

Then, by (2)
`e^2 = e^(ln(x^2+x)) = x^2 + x`
`=> x^2 + x - e^2 = 0`

Now we apply the quadratic formula with `a=b=1` and `c=-e^2`

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

But as noted at the beginning, we must have `x > 0` and so we reject `(-1-sqrt(1+e^2))/2`

Thus we are left with the answer

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