# Math Help - Tricky (ln) problem

1. ## Tricky (ln) problem

I was assigned the following:

The only solution of the equation ln(x) + ln(x-2) =1 is x = ?

I am unsure how to solve for x.

2. ## Re: Tricky (ln) problem

Use the fact that $\ln(a)+\ln(b)=\ln(a\cdot b)$ and $\ln(e)=1$ therefore the equation can be arranged as:
$\ln[x(x-2)]=\ln(e)$
$\Leftrightarrow x(x-2)=e$

3. ## Re: Tricky (ln) problem

I solved for x, and came up with 1 + sqrt(4 +4e). However, this is still wrong, and I am unsure why!

4. ## Re: Tricky (ln) problem

Originally Posted by Dmiller93
I solved for x, and came up with 1 + sqrt(4 +4e). However, this is still wrong, and I am unsure why!
Following on from Siron's working:

$x^2-2x = e \Leftrightarrow x^2-2x-e = 0$

Since e is a number solve using the quadratic formula:

$x = \dfrac{2\pm \sqrt{4+4e}}{2} = \dfrac{2 \pm \sqrt{4(1+e)}}{2} = \dfrac{2 \pm \sqrt{4}\sqrt{1+e}}{2}$

$= \dfrac{2 \pm 2\sqrt{1+e}}{2} = \dfrac{2(1 \pm \sqrt{1+e})}{2} =1 \pm \sqrt{1+e}$

Since $\sqrt{1+e} > 1$ discard the negative solution due to domain issues (x>2)

$x = 1 + \sqrt{1+e}$