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.

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- Sep 23rd 2011, 12:23 PMDmiller93Tricky (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. - Sep 23rd 2011, 12:27 PMSironRe: Tricky (ln) problem
Use the fact that $\displaystyle \ln(a)+\ln(b)=\ln(a\cdot b)$ and $\displaystyle \ln(e)=1$ therefore the equation can be arranged as:

$\displaystyle \ln[x(x-2)]=\ln(e)$

$\displaystyle \Leftrightarrow x(x-2)=e$

Solve this quadratic equation. - Sep 23rd 2011, 12:49 PMDmiller93Re: 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!

- Sep 23rd 2011, 01:03 PMe^(i*pi)Re: Tricky (ln) problem
Following on from Siron's working:

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

Since e is a number solve using the quadratic formula:

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

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

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

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