This is a typical differential equation problem. It leads to what is known as a "logistics model".
Let $N(t)$ denote the number of infected people at time $t$ after the first infection, with $t$ measured in days. The rate at which the disease spreads is the rate of change of $N(t)$, which is the derivative $N'(t)$. We are told that the rate of change is proportional to $N(t)$ (the number of people who already have the disease) times $10000-N(t)$ (the number of people who don't have the disease). That is:
$$N'(t) = kN(t)(10000-N(t))$$
for some constant $k$.
We are also told that $N(0) = 1$ (one person gets sick) and $N(5)=50$. We are asked for the value of $N(10)$.
Note that we cannot have $N(t)=0$ for all $t$, nor can we have $N(t)=10000$ for all $t$.
This differential equation is of a kind called "separable". We can solve it as follows:
$$\begin{align*}
\frac{dN}{dt} &= kN(10000-N)\\
\frac{dN}{N(10000-N)} &= k\,dt\\
\int\frac{dN}{N(10000-N)} &= \int k\,dt\\
\int\frac{dN}{N(10000-N)} &= kt + C.
\end{align*}$$
for some constant $C$. To solve the integral on the left, we can use Partial Fractions:
$$\frac{1}{N(10000-N)} = \frac{\quad\frac{1}{10000}\quad}{N} + \frac{\quad\frac{1}{10000}\quad}{10000-N}$$
so
$$\begin{align*}
\int\frac{dN}{N(10000-N)} &= \frac{1}{10000}\int\left(\frac{1}{N} + \frac{1}{10000-N}\right)\,dN\\
&= \frac{1}{10000}\left(\ln|N| - \ln|10000-N|\right) + D.
\end{align*}$$
Putting it all together, we get
$$\frac{1}{10000}\left(\ln |N| - \ln|10000-N|\right) = kt + E,$$
where $E$ is some constant. Rewriting, we have:
$$\ln\left|\frac{N}{10000-N}\right| = Kt + F,$$
where $K$ and $F$ are constants. Since $N$ is always between $1$ and $10000$, we can drop the absolute values.
Plugging in $t=0$, we know that $N=1$, so we have
$\ln\frac{1}{9999} = F$. Plugging in $N=5$, we know that $N(t)=50$, so we have
$$\ln\frac{50}{9950} = 50K + \ln\frac{1}{9999}.$$
From here, we can plug in $t=10$ and solve for $N$.
