Solution
- We need to find the expression for the population and the way to do that is to integrate both sides with respect to time.
- After this, we can then apply the values given to solve the question.
- This is done below:
[tex]\begin{gathered} \frac{dP}{dt}=kP \\ \\ \text{ Rearrange} \\ \frac{dP}{P}=kdt \\ \\ \text{ Integrate both sides} \\ \int\frac{dP}{P}=\int kdt \\ \\ \ln P=kt+C \\ \text{ Take the }e\text{ exponent of both sides} \\ \\ e^{\ln P}=e^{kt+C} \\ \\ \text{ But we know that }e^{\ln x}=x \\ \\ \text{ Thus, we have that:} \\ P=e^{kt+C}=e^{kt}.e^C \\ \text{ Let }A=e^C,\text{ since }e^C\text{ is a constant} \\ \\ \therefore P=Ae^{kt} \\ \\ \\ \\ Now\text{ that we have the expression for P, let us find the values of }A,\text{ and }k\text{ using the values} \\ given. \\ \\ \text{ when }P=40000,t=3: \\ 40000=Ae^{3k} \\ when\text{ }P=50000,t=5: \\ 50000=Ae^{5k} \\ \\ \text{ Divide both equations} \\ \frac{50000}{40000}=\frac{Ae^{5k}}{Ae^{3k}} \\ \\ \frac{5}{4}=e^{5k-3k}=e^{2k} \\ \\ \text{ Take the natural log of both sides} \\ \ln\frac{5}{4}=\ln e^{2k}=2k \\ \\ \therefore k=\frac{1}{2}\ln(\frac{5}{4})=\ln\frac{\sqrt{5}}{2} \\ \\ \text{ Putting this value of }k\text{ into any of the two equations,} \\ 40000=Ae^{\ln\frac{\sqrt{5}}{2}\times3}=Ae^{\ln(\frac{\sqrt{5}}{2})^3} \\ 40000=A\times(\frac{\sqrt{5}}{2})^3=A\times\frac{5\sqrt{5}}{8} \\ \\ A=\frac{40000\times8}{5\sqrt{5}}=\frac{64,000}{\sqrt{5}}\times\frac{\sqrt{5}}{\sqrt{5}}=12,800\sqrt{5} \\ \\ \\ \text{ Thus, our population equation is:} \\ P=12,800\sqrt{5}e^{\ln(\frac{\sqrt{5}}{2})^t} \\ P=12,800\sqrt{5}(\frac{\sqrt{5}}{2})^t \end{gathered}[/tex]- Now that we have the expression for P, we can thus evaluate it at t = 10 as follows:
[tex]\begin{gathered} P=12,800\sqrt{5}(\frac{\sqrt{5}}{2})^{10} \\ \\ P=87,346.40537...\approx87,346\text{ \lparen To the nearest integer\rparen} \end{gathered}[/tex]Final Answer
The population is 87,346
