Differential Equation Disease Modelling 
The rate at which people are infected per week after the outbreak of a disease is $${dP\over dt} = 0.8P \left(1 - {P\over 350}\right)$$ At the outbreak for the disease, $7$ people were infected. What is the initial condition problem for $P$?

I got everything to the left side with dP and left only 1 along with dt on the right. Then, using partial fractions, I got 
∫${1\over 0.8P}$ + ${1\over 280(1-{P\over 350})}$ 
I integrated this further to get 
${-5\over 4}$ln(350-P) + ${5\over 4}$ln(P)
I then did  ${ln(P)^{5/4}\over (350-P)^{5/4}}$ = t + c
I was unsure of what to do after this and how I can substitute the initial conditions (P as 7 and t as 0?, I'm not so sure). 

I think that your last formula is not correct.
You correctly wrote
$$\frac{5 }{4}\log (P)-\frac{5}{4} \log (350-P)=t+c$$ which rewrite
$$\frac{5 }{4}\log \left(\frac{P}{350-P}\right)=t+c_1\implies \log \left(\frac{P}{350-P}\right)=\frac{4 }{5}t+c_2$$ Now, exponentiate both sides
$$\frac{P}{350-P}= \exp\left(\frac{4 }{5}t+c_2 \right)=c_3\exp\left(\frac{4 }{5}t \right)$$ Now, solve for $P$
$$P=\frac{350\, c_3\, e^{4 t/5}}{1+c_3\, e^{4 t/5}}$$ Now, use the condition
$$7=\frac{350\, c_3}{1+c_3}\implies c_3=\frac 1{49}\implies P=\frac{350 \,e^{4 t/5}}{49+e^{4 t/5}}$$
