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.

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.