Let the number of infected students after $t$ days be $x(t)$.

Total students $=100$, so non-infected students $=100-x$.

Given rate of spread is directly proportional to the product of infected and non-infected students: $\frac{dx}{dt}=k\,x(100-x)$ where $k>0$ is a constant.

Separate variables and integrate.

$\frac{dx}{x(100-x)}=k\,dt$

Using partial fractions, $\frac{1}{x(100-x)}=\frac{1}{100}\left(\frac{1}{x}+\frac{1}{100-x}\right)$

So, $\int \frac{dx}{x(100-x)}=\frac{1}{100}\int \left(\frac{1}{x}+\frac{1}{100-x}\right)dx$

Hence, $\frac{1}{100}\left(\ln x-\ln(100-x)\right)=kt+C$

or $\ln\left(\frac{x}{100-x}\right)=100kt+C_1$

Therefore, $\frac{x}{100-x}=Ae^{100kt}$ for some constant $A$.

Use initial condition $x(0)=2$.

$\frac{2}{100-2}=A$ $A=\frac{2}{98}=\frac{1}{49}$

So, $\frac{x}{100-x}=\frac{1}{49}e^{100kt}$

Use the condition $x(4)=30$.

$\frac{30}{70}=\frac{1}{49}e^{400k}$ $\frac{3}{7}=\frac{1}{49}e^{400k}$ $e^{400k}=49\cdot \frac{3}{7}=21$

Find $x(8)$.

$\frac{x(8)}{100-x(8)}=\frac{1}{49}e^{800k}$

Since $e^{400k}=21$, $e^{800k}=21^2=441$

Thus, $\frac{x(8)}{100-x(8)}=\frac{441}{49}=9$

So, $x(8)=9(100-x(8))$ $x(8)=900-9x(8)$ $10x(8)=900$ $x(8)=90$

Final Answer

Number of infected students on the 8th day is: $\boxed{90}$