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
asked 2020-12-13
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

Answers (1)

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.
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