Answer:
(1) Let, P represents the size of the bacteria culture in t hours,
According to the question,
[tex]\frac{dP}{dt}\propto P[/tex]
[tex]\frac{dP}{dt}=kP[/tex]
[tex]\frac{dP}{P}=kdt[/tex]
By integrating,
[tex]\ln P=kt + C[/tex]
[tex]P=e^{kt+C}[/tex]
[tex]P=e^{kt}.e^C[/tex]
[tex]P=P_0 e^{kt}[/tex] Where [tex]e^C=P_0[/tex],
We have,
at t = 0, P = 200,
[tex]\implies 200=P_0 e^0\implies P_0 = 200[/tex]
at t = 6, P = 1200
[tex]1200=200 e^{6k}\implies k = 0.299[/tex]
Hence, the required function would be,
[tex]P=200 e^{0.299t}[/tex]
(2) if t = 7,
The population would be,
[tex]P=200 e^{0.299\times 7}=1621.84126567\approx 1622[/tex]
(3) If P = 1750,
[tex]1750=200 e^{0.299t}\implies t = 7.254[/tex]
Hence, it will take about 8 years for the population to reach 1750.