If the daily increase in the population is p then 100p²=300 between days 2 and 4.
Therefore p²=3 and p=√3. We can write p(t)=p₀3^(t/2) where p₀ is the initial population.
After 6 days, which is 2 more than 4 days, there will be 300p²=900 flies.
So since p₂=3p₀, p₀=p₂/3=100/3 and p(t)=3^(t/2)×100/3 is a model for the situation.
5000×3/100=3^(t/2); 3^(t/2)=5000×0.03=150, (t/2)log(3)=log(150) from which t=2log(150)/log(3)=9.12 approx.
So after 10 days there will be more than 5000 flies.