# Linearizing

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• Sep 1st 2008, 06:59 PM
Chayned
Linearizing
Im in need of some serious help.

I'm given the problem of the theoretical radioactive decay.

N=(N0)(e^(-.693t/t½))

It needs to look like y=mx+b
t½= t sub 1/2
N0= N sub zero
y=N/No

This is linearizing equations and I dont even know where to begin. Any help at all?
• Sep 2nd 2008, 03:00 AM
Sean12345
Hi Chayned,

Let's define $\displaystyle \lambda = t_{\frac{1}{2}}$ . We have

$\displaystyle N=N_0\cdot e^{\left(\dfrac{-0.693t}{\lambda}\right)}$

Well a good start would be to log both sides with base e (You will see why in a minute)

$\displaystyle \implies \ln(N)=\ln \left\{N_0\cdot e^{\left(\dfrac{-0.693t}{\lambda}\right)}\right\}$

$\displaystyle \implies \ln(N)=\ln(N_0)+\ln \left\{e^{\left(\dfrac{-0.693t}{\lambda}\right)}\right\}$ Since $\displaystyle \ln (ab) \equiv \ln (a) + \ln (b)$

$\displaystyle \implies \ln (N)=\frac{-0.693t}{\lambda}+ \ln (N_0)$ Since $\displaystyle \ln (e^a) \equiv a$

$\displaystyle \implies \ln (N)=\frac{-0.693}{\lambda} \cdot t + \ln (N_0)$ Which is of the form $\displaystyle y=mx+b$ where $\displaystyle y=\ln (N)~,~m=\frac{-0.693}{\lambda}~,~x=t$ and $\displaystyle b=\ln (N_0)$