# Thread: [SOLVED] solving a log

1. ## [SOLVED] solving a log

I'm having difficulties solving this log. I was hoping someone could point me in the right direction
$\displaystyle lnx +ln(x+4) = 2$
$\displaystyle lx(x(x+4)) = 2$
$\displaystyle x(x+4) = e^2$

I could write that out to a sorta-polynomial of
$\displaystyle x^2 + x - e^2$
but I'm not really sure where I should be proceeding from here. Thanks in advance for any assistance.

2. Use the Quadratic Formula with $\displaystyle a = 1$, $\displaystyle b = 1$, and $\displaystyle c = -e^2$.

3. I'd tried that route by my answer was way off from what the book had. Let me do the work and perhaps you could point out what I'm doing wrong?

$\displaystyle x^2+x-e^2$
$\displaystyle \frac{-1 + \sqrt{1+4(e^2)}}{2}$
$\displaystyle \frac{-1 + \sqrt{5e^2}}{2}$

If I solve that I get a positive answer of 2.539 (roughly) and a negative answer which is invalid. Unfortunately the book has

$\displaystyle 1 + \sqrt{1+e} \approx 2.928$

4. I believe the error lies in your conversion of $\displaystyle x(x + 4) = e^2$ to $\displaystyle x^2 + x - e^2 = 0$, specifically your distribution of x to 4.

5. Originally Posted by satis
I'm having difficulties solving this log. I was hoping someone could point me in the right direction
$\displaystyle lnx +ln(x+4) = 2$
$\displaystyle lx(x(x+4)) = 2$
$\displaystyle x(x+4) = e^2$

I could write that out to a sorta-polynomial of
$\displaystyle x^2 + x - e^2$
but I'm not really sure where I should be proceeding from here. Thanks in advance for any assistance.
You distributed $\displaystyle x$ incorrectly in that last step, try using the quadratic formula on:
$\displaystyle x^2+4x-e^2$
Or, you could factor $\displaystyle x(x+4)-e^2=0$ fairly easily.

6. sure enough... for some reason I kept reading that 4 as a 1. Repeatedly. Thanks for pointing out the issue guys.