Solving for x with ln and e

[quicknote]

The question is: for a Gaussian distribution what is the mathematical relationship between the FWHM and the standard deviation.

The equations I'm using are:

N(x) =\frac {\ a}{2}
N(x) = Ae^- \frac {\ (x-x_2)^2}{2 sigma^2}

I equated the equations and started to solve for x. I know you take the ln of both sides to get rid of e. But a few of my friends have 2ln2 in their answer...how do you get ln2? isn't it ln 1/2?

[Tide]

The "extra" factor of 2 arises from the fact that there are two values of x for which N = A/2 and the difference between them gives you the FWHM.

[quicknote]

Sorry, I mean the 2 in ln2, not the factor 2

N(x) = A/2 ....the A cancels, giving you 1/2
then taking the ln of both sides would give you ln 1/2, right? not ln2

I'm getting [tex]sqrt(2*sigma*ln \frac {\1}{2}) = -(x-x_2)


Thanks.

[Tide]

You do understand that \ln \frac {1}{2} = - \ln 2?