Answer:
The population of mosquitoes in the area at any time <em>t</em> is:

Step-by-step explanation:
The rate of growth of mosquitoes can be expressed as:

Integrate the above expression as follows:


It is provided that the population doubles every day.
Compute the value of <em>k</em> as follows:

It is also provided that every day 20,000 mosquitoes are eaten.
The rate of growth per week can be expressed as:

The integrating factor for this is:

Then,
![P(t)\ e^{-\ln(2)t}=\int {e^{-\ln(2)t}}-14000\, dt\\=-14000\int {e^{-\ln(2)t}}\, dt\\=-14000\times \frac{e^{-\ln(2)t}}{-\ln(2)}+C\\P(t)=(e^{-\ln(2)t})\times [-14000\times \frac{e^{-\ln(2)t}}{-\ln(2)}+C]\\=\frac{14000}{\ln(2)}+Ce^{-\ln(2)t}](https://tex.z-dn.net/?f=P%28t%29%5C%20e%5E%7B-%5Cln%282%29t%7D%3D%5Cint%20%7Be%5E%7B-%5Cln%282%29t%7D%7D-14000%5C%2C%20dt%5C%5C%3D-14000%5Cint%20%7Be%5E%7B-%5Cln%282%29t%7D%7D%5C%2C%20dt%5C%5C%3D-14000%5Ctimes%20%5Cfrac%7Be%5E%7B-%5Cln%282%29t%7D%7D%7B-%5Cln%282%29%7D%2BC%5C%5CP%28t%29%3D%28e%5E%7B-%5Cln%282%29t%7D%29%5Ctimes%20%5B-14000%5Ctimes%20%5Cfrac%7Be%5E%7B-%5Cln%282%29t%7D%7D%7B-%5Cln%282%29%7D%2BC%5D%5C%5C%3D%5Cfrac%7B14000%7D%7B%5Cln%282%29%7D%2BCe%5E%7B-%5Cln%282%29t%7D)
The initial population is 200,000.
Compute the value of <em>C</em> as follows:

Now substitute <em>C</em> in P (t),
