[SOLVED] solving a log

**satis** · February 22nd 2010, 03:55 PM

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

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

**NOX Andrew** · February 22nd 2010, 04:00 PM

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

**satis** · February 22nd 2010, 04:09 PM

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?

$x^2+x-e^2$
into the quadratic equation
$\frac{-1 + \sqrt{1+4(e^2)}}{2}$
$\frac{-1 + \sqrt{5e^2}}{2}$

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

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

**NOX Andrew** · February 22nd 2010, 04:28 PM

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

**Dotdash13** · February 22nd 2010, 05:20 PM

Originally Posted by satis

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

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

You distributed $x$ incorrectly in that last step, try using the quadratic formula on:
$x^2+4x-e^2$
Or, you could factor $x(x+4)-e^2=0$ fairly easily.

**satis** · February 22nd 2010, 06:28 PM

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

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