$S(t)=S_{0}e^{-R_{0}(R(t)-R_{0})/N}$ 
Show that from the system of equations generated by the SIR model we have $$S(t) = S(0) \exp\left(-R_{0}(R(t)-R(0))/N \right)$$ where the model SIR equations are as follows $$\begin{aligned} \frac{dS}{dt} &= -\beta \frac{SI}{N}\\ \frac{dI}{dt} &=  \beta \frac{SI}{N}-\gamma I\\ \frac{dR}{dt} &= \gamma I \end{aligned}$$ where $t \geq 0^{+}$.

 A: Sort of an odd problem, but here goes. If $S$, $I$, and $R$ satisfy this system, your first equation tells you that $S'/S = -(\beta/N)I$. So
\begin{align}
S'/S &= -\frac{\beta}{N}I \\
&= -\frac{\beta}{N\gamma} R',
\end{align}
the last line following from your third relation.
And you know that $S'/S = (\log S)'$, so
$(\log S)' = -(\beta/N\gamma)R'$ which tells you
$$ \log S = C -(\beta/N\gamma)R.$$
where $C$ is a constant of integration. Knowing the initial conditions for $S$ and $R$ let you figure out what $C$ is, and you get
$$
S = S_0 \exp(\frac{\beta}{\gamma N}(R_0 - R)),$$
which probably is what you meant to type in the title.
I imagine in practice you wouldn't know $R$ beforehand. Least squares estimation is one way to get a handle on the parameters if you happen to have measurements. Let's suppose you have measurements of $S$ and $I$:
$(s_1, i_1), (s_2, i_2), \dots$. You might set up some sort of error functional
\begin{align}
E(s(0), i(0), \beta/N, \gamma)
=& \sum_n (S(t_n; s(0), i(0), \beta/N, \gamma) - s_n)^2 \\
&+ \sum_n(I(t_n; s(0), i(0), \beta/N, \gamma) - i_n)^2
\end{align}
and adjust the parameters
$s(0)$, $i(0)$, $\beta/N$, and $\gamma$ until you get the accuracy you want, solving the system numerically to generate the associated $S$ and $I$ each time you change them, or better yet take the advice of @CarlChristian and use the exact solution!
