# Find an exponential function of the form P(t)=P_0n^{frac{t}{T}} that models the situation, and then find the equivalent exponential model of the form P(t)=P_0e^{rt} 1) Doubling time of 25 weeks, initial population of 1300

Question
Exponential models
Find an exponential function of the form $$P(t)=P_0n^{\frac{t}{T}}$$ that models the situation, and then find the equivalent exponential model of the form $$P(t)=P_0e^{rt}$$
1) Doubling time of 25 weeks, initial population of 1300

2020-12-14
Given: Initial population $$=P_0=1300$$ at t=0
Population $$=2\timesP_0$$ at t=26 week
Use the given model
$$P(t)=P_0n^{\frac{t}{T}}$$
Substitute $$P_0=1300$$
$$P(t)=1300n^{\frac{t}{T}}$$
Substitute $$P(t)=2\times1300$$ and $$t=25$$ weeks
$$2\times1300=1300n^{\frac{25}{T}}$$
$$n^{\frac{25}{T}}=\frac{2\times1300}{1300}$$
$$n^{\frac{25}{T}}=2$$
This equation holds true only if n=2 and T=25
Therefore, the possible model for the given situation is $$P(t)=1300(2)^{\frac{t}{25}}$$ where t is in weeks and P(t) is the population at any time t.
Since
The equation $$P(t)=P_0n^{\frac{t}{T}}$$ is equivalent to the equation $$P(t)=P_0e^{et}$$ thherefore
$$P_0n^{\frac{t}{T}}=P_0e^{rt}$$
$$n^{\frac{t}{T}}=e^{rt}$$
Take ln both sides of the equation
$$\ln(n)^{\frac{t}{T}}=\ln e^{rt}$$
$$\frac{t}{T}\ln(n)=rt$$
$$\frac{1}{T}\ln(n)=r$$
Here, substitute the value if n=2 and T=25
$$\frac{1}{25}\ln(2)=r$$
$$r=0.0277$$
Hence, the required model is $$P(t)=1300e^{0.0277t}$$ where t is in weeks.

### Relevant Questions

Find an exponential function of the form $$\displaystyle{P}{\left({t}\right)}={P}_{{0}}{n}^{{{\frac{{{t}}}{{{T}}}}}}$$ that models the situation, and then find the equivalent exponential model of the form $$\displaystyle{P}{\left({t}\right)}={P}_{{0}}{e}^{{{r}{t}}}$$
1) Doubling time of 25 weeks, initial population of 1300
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(e) Explain why the exponential model is a good approximation to the logistic model when $$\displaystyle\frac{{N}}{{K}}$$ is small compared with 1.
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