Exponential decay word problem

How would I solve the following word problem.

A radioactive substance weighed $n$ grams at time t=0. Today 5 years later the substance weight $m$ grams.How much will the substance weight 5 years from now?

How would I solve this I know the formula I have to use is $f(x)=Ce^{kt}$ and to find $k$ you do $\ln2$/half life

• Hint, in equal periods of time, the exponential decay is same (expressed as a ratio). That's why you say there's a half life. Commented Sep 4, 2013 at 17:35

We don't need to calculate $k$. We have $$f(t)=Ce^{kt}.$$

We are told that $f(0)=n$, so $C=n$.

We are told that $f(5)=m$. So $$Ce^{5k}=ne^{5k}=m.\tag{1}$$

We want to find $f(10)$, which is $ne^{10k}$. But from (1) $e^{5k}=\frac{m}{n}$, so $e^{10k}=\frac{m^2}{n^2}$ and therefore $$f(10)=n\cdot \frac{m^2}{n^2}=\frac{m^2}{n}.$$

• I think I understand the solution but one thing I do not get is why from $e^5k=m/n$ it is squared to $e^{10k}=m^2/n^2$ Commented Sep 4, 2013 at 17:45
• In general, $(e^a)^2=e^{2a}$. Special case of the probably familiar formula $(a^b)^c=a^{bc}$. Commented Sep 4, 2013 at 17:54

Hint:

What does $f(0) = Ce^{0} = n$ tell you about $C$?

What does $f(5) = Ce^{5k} = m$ tell you about $k$?

What do you then get when you calculate $f(10)$?

• Hmm does $f(0)=Ce^{(0)}$ tell me C is zero Commented Sep 4, 2013 at 17:33
• No, $e^0=1$, so $C =$ ? Commented Sep 4, 2013 at 17:34
• C would be 1 I think also k would be 1 Commented Sep 4, 2013 at 17:35
• No, take a look at the expression $Ce^0=n$ again. If $e^0=1$, then what's left? Commented Sep 4, 2013 at 17:36
• n would be left I see Commented Sep 4, 2013 at 17:40

The function that gives the radioactive substance weight is

$W(t)=n*e^{-ct}$ ($n$=weight at time $t=0$)

We have

$W(5)=n*e^{-5c} \implies m=n*e^{-5c}$

$e^{-5c}=m/n \implies -5c=ln(\frac{m}{n})\implies c=\frac{-1}{5}ln(\frac{m}{n})$.

So the function becomes

$W(t)=n*e^{\frac{1}{5}ln(\frac{m}{n})t}$

Finally after $5$ years from now

($10$ years from $t=0$)

$W(10)=n*e^{2ln(\frac{m}{n})}= n*e^{{ln(\frac{m}{n})^2}}=n*(\frac{m}{n})^2=\boxed{\frac{m^2}{n}}$