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### Half-Life of a Radioactive Substance

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Date: 4 Aug 1995 17:10:44 -0400
From: Margie Salaz
Subject: Half-Life

Dear Ken,

Here is my question:  The half-life of a radioactive substance is d days.
If you begin with a sample weighing b g, how long will it be until c g
remains?

Here is what I did:  I used the Decay Formula: N = No e^kt
1/2No = No e^dk
1/2 = e^dk
(1/2)^1/d = (e^dk)^1/d = ek
(1/2)^1/d = ek

cg = bg (e^k)^t
Cg/bg = bg/bg <(1/2)^1/d>^t
c/b = (1/2)^1/d^t

Here is where I began to get bogged down. Will you please see what I
have done wrong here:

c/b = (1/1/2^d)^t
d c/b = 1/1/2^t
d In c/b/In 1/2 = t

The last line is the right answer, but I think my last 3 steps aren't in
line with the final product.

Will you help me?

Thanks,

Margie Salaz
Cuba, NM
```

```
Date: 7 Aug 1995 09:29:01 -0400
From: Dr. Ethan
Subject: Re: Half-Life

Well I am quite impressed - you made my job very easy.  You knew exactly
where your mistake was.  I will start there.

>cg = bg (e^k)^t
>Cg/bg = bg/bg <(1/2)^1/d>^t
>c/b = (1/2)^1/d^t

Up to here you are fine.  But here is where you make your error.  Your
next line would be correct if it was -d not 1/d.  What you should do next
is raise both sides to the d power.

Then you have

(c/b)^d = (1/2)^t

Then take the ln (natural log) of both sides: you get

ln [ (c/b)^d ] = ln [ (1/2)^t]

which equals  d ln[c/b] = t ln[1/2]

So t= d * ln[c/b]/ln[1/2]

Hope that helps.

Ethan Doctor On Call
```
Associated Topics:
High School Calculus

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