So you need to solve

\[

\frac{dN}{dt} = k N \ln(\frac{M}{N})

\]

for $N$. The authors say that you should make the substituion $u = ln(M/N)$. I think it’s easier to start by realizing that $ln(M/N) = -ln(N/M)$ (try it for some positive values of $M$ and $N$ on your calculator if you don’t believe me). So we have:

\[

\frac{dN}{dt} = -k N \ln(\frac{N}{M})

\]

Now letting $u = \ln(\frac{N}{M})$, we can rewrite this as

\[

\frac{dN}{dt} = -k N u

\]

This still sucks, because we seem to have two variables ($N$ and $u$) in our formula. But they’re not really two variables, because of course $u = ln(N/M)$, where $M$ is a constant. So the only calculus trick is to realize that by the Chain Rule, and the fact that $\frac{d}{dx}\ln(x) = \frac{1}{x}$ we

can write

\[

\frac{du}{dt} = \frac{1}{N} \frac{dN}{dt}

\]

So this allows us to make a sequence of steps that eliminate $N$ and convert the equation to be in terms of $u$,

which is the answer to part b:

\[

\frac{du}{dt} = \frac{1}{N} \frac{dN}{dt}

\]

\[

\frac{dN}{dt} = N \frac{du}{dt}

\]

\[

N \frac{du}{dt} = -k N u

\]

\[

\frac{du}{dt} = -k u

\]

Does this last equation (answer to part b) look familiar? Of course it does: it’s the formula for unconstrained decay, and we know the solution is

\[

u = u_0 e^{-kt}

\]

which is the answer to part c.

To complete the exercise, we replace $u$ in the previous equation with its original form $\ln(\frac{N}{M})$:

\[

\ln(\frac{N}{M}) = \ln(\frac{N_0}{M}) e^{-kt}

\]

So now do some algebra tricks, and you’ll get the answer in terms of $N$: in other words, the answer to parts (d) and (e), which are really not two questions but one, as far as I can tell.