The question is as follows:

I have the general solution of a differential equation as $P(t) = Ae^{kt}$ where $k < 0$ is a constant, and it describes the mass $P(t)$ of a radioactive isotope at future time $t$. I also have that the mass of the isotope is $10$ at time $0$ and it is $5$ at time $2$. I want to find the remainder of the isotope at time $3$.

So I can use that $P(0) = 10$ to obtain $A$ = $10$, and here now is where I get a bit confused. If I use $P(2) = 5$, I get $5 = 10e^{2k}$ and solving for $k$ I get $k = \frac{ln(0.5)}{2} = -0.34657$. And then substitute it all back in to get $P(3)$

However, when looking at other questions online, they talk a lot about half-life of the substance and we have that if we let $\tau$ denote the half-life of the substance, then $k = \frac{ln(2)}{\tau}$ (where they are using the general solution to be $Ae^{-kt}$ However, isn't $\tau$ in my case equal to $2$? So should I not be getting $k = \frac{ln(2)}{2}$ rather than $k = \frac{ln(0.5)}{2}$? Can someone please explain where I am going wrong?

Thank you

• You haven't don't anything wrong because $-\log{(0.5)}=\log{(2)}$. Which explains the negative sign, your solution is the same as theirs Mar 20 '18 at 12:51
• Ah ok I think I get where I am getting confused. So what I have written is fine, I just solved for $k$ where $k$ is negative whereas the one online solved for $k$ where $k$ is positive and when putting back into their formula they will take this into account and multiply it by negative $1$? Mar 20 '18 at 12:55

$Ae^{-kt}$ with $k=\frac{ln(2)}{\tau}$ is equivalent to $Ae^{kt}$ with $k=\frac{ln(0.5)}{\tau}$
• Thanks very much. One last thing, so using that then $P(3)$ is approx equal to $3.5$ yes? Mar 20 '18 at 12:58
• Yes, it makes sense as you can double check with half-life equal to two, which means that $P(4)=2.5$, and you know the exponential decay function is decreasing. Mar 20 '18 at 13:00