Question

this is the only question in this section. question the logistic model can be written as: $p(t)=\frac{p_0\cdot k}{p_0+(k - p_0)\cdot e^{-rt}}$ or $p(t)=\frac{k}{1+\frac{(k - p_0)}{p_0}e^{-rt}}$ $p(t)$ is population size at time $t$, $p_0$ is the initial population size, $k$ is the carrying capacity of the environment (maximum population size an environment can support), and $r$ is a constant representing the rate of the population growth or decay. exercise 1: the population of sasquatch in bigfoot county is modeled by $p(t)=\frac{120}{1 + 4e^{-0.05t}}$ where $p(t)$ is the population of sasquatch $t$ years after 2010. when will the population of sasquatch in bigfoot county reach 60? give the exact answer(without calculator).

Explanation:

Step1: Substitute \( P(t) = 60 \) into the formula

Given \( P(t)=\frac{120}{1 + 4e^{-0.05t}}\), substitute \( P(t)=60 \). So, \( 60=\frac{120}{1 + 4e^{-0.05t}}\).

Step2: Cross - multiply

Cross - multiply the equation \( 60=\frac{120}{1 + 4e^{-0.05t}}\) to get \( 60(1 + 4e^{-0.05t})=120 \). Then divide both sides by 60: \( 1 + 4e^{-0.05t}=2 \).

Step3: Solve for \( e^{-0.05t} \)

Subtract 1 from both sides of \( 1 + 4e^{-0.05t}=2 \), we have \( 4e^{-0.05t}=1 \). Then divide both sides by 4: \( e^{-0.05t}=\frac{1}{4} \).

Step4: Take the natural logarithm of both sides

Since \( y = e^{x}\) and \( x=\ln(y)\) are inverse functions, for \( e^{-0.05t}=\frac{1}{4} \), we take the natural logarithm of both sides. \(\ln(e^{-0.05t})=\ln(\frac{1}{4})\).
Using the property \(\ln(e^{a})=a\), the left - hand side is \(-0.05t\). And \(\ln(\frac{1}{4})=\ln(4^{-1})=-\ln(4)\).

Step5: Solve for \( t \)

We have \(-0.05t=-\ln(4)\). Multiply both sides by \(- 1\): \(0.05t=\ln(4)\). Then \(t=\frac{\ln(4)}{0.05}\). Since \(4 = 2^{2}\), \(\ln(4)=\ln(2^{2}) = 2\ln(2)\). So \(t=\frac{2\ln(2)}{0.05}=40\ln(2)\).

Answer:

\(t = 40\ln(2)\) years after 2010.