$\begin{array}{}b(t)=\frac{c}{1+A{e}^{-Bt}},\text{}where\text{}c\text{}is\text{}canuing\text{}capacity\\ and-B=\frac{1}{{t}_{1}}\mathrm{ln}(\frac{{P}_{0}({P}_{5}-{P}_{1})}{{P}_{1}({P}_{5}-{P}_{0})})\text{}and\text{}A=\frac{c-{P}_{0}}{{P}_{0}}\\ Given\text{}that,\text{}at\text{}t=0,{P}_{0}=300\\ t=1,{P}_{1}=700\\ {P}_{5}=5400\\ As\text{}P(t)=\frac{c}{1+A{e}^{-Bt}},\text{}here\text{}c=5400\\ And\text{}-B=\frac{1}{{f}_{1}}\mathrm{ln}(\frac{{P}_{0}({P}_{5}-{P}_{1})}{{P}_{1}({P}_{5}-{P}_{0})})\\ =\frac{1}{1}\mathrm{ln}(\frac{300(5400-700)}{700(5400-300)})\\ =\mathrm{ln}(\frac{47}{119})\\ A=\frac{5400-300}{300}=\frac{5100}{300}=17\\ =\frac{5400}{1+\frac{47}{7}}t\end{array}$