Recent content by SReinhardt

So if you used based base .5 instead of base e, you'd get what I worked out on my own. The main thing that would change then would be the \lambda

Wouldn't there have to be a negative sign in there somewhere >_> I believe you have to use integrals to solve that, which I haven't done yet.

I've never had any issues using it...I can see where you're coming from though. Your point is for when you're solving for the time or half-life. But then all you have to do is take the log10 and divide. It could also be I use it just to make my teacher grade things two ways xD

I'm curious on why they chose to use N=N_0 e^{- \lambda t} instead of N = N_O .5^{\frac{t}{half-life}} The 2nd one is one that I figured out, and it makes more sense to me; it is based off the idea of half-lives. (I'm not saying it's original or hasn't been done before, just was never shown to me)

What lead them to use e and the natural log of 2 in the decay formula? A much simpler (to me at least) method would is: N=No*.5^(time/half life)