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.