if the population of squirrels on campus t years after the beginning of 1855 is given by the logistic growth…

if the population of squirrels on campus t years after the beginning of 1855 is given by the logistic growth function s(t)=5000 / (1 + 18e^(-0.43t)). find the time t such that s(t)=3750. time (exact value without using a calculator), t = ln(54) / 0.43 which corresponds to the year (rounded to two decimal places)

if the population of squirrels on campus t years after the beginning of 1855 is given by the logistic growth function s(t)=5000 / (1 + 18e^(-0.43t)). find the time t such that s(t)=3750. time (exact value without using a calculator), t = ln(54) / 0.43 which corresponds to the year (rounded to two decimal places)

Answer

Explanation:

Step1: Set up the equation

Set $s(t)=3750$ in the logistic - growth function $s(t)=\frac{5000}{1 + 18e^{-0.43t}}$, so we have $3750=\frac{5000}{1 + 18e^{-0.43t}}$.

Step2: Cross - multiply

Cross - multiplying gives $3750(1 + 18e^{-0.43t})=5000$. Then $1 + 18e^{-0.43t}=\frac{5000}{3750}=\frac{4}{3}$.

Step3: Isolate the exponential term

Subtract 1 from both sides: $18e^{-0.43t}=\frac{4}{3}-1=\frac{1}{3}$. Then $e^{-0.43t}=\frac{1}{3\times18}=\frac{1}{54}$.

Step4: Take the natural logarithm

Taking the natural logarithm of both sides, $\ln(e^{-0.43t})=\ln(\frac{1}{54})$. Since $\ln(e^{-0.43t})=-0.43t$ and $\ln(\frac{1}{54})=-\ln(54)$, we get $-0.43t=-\ln(54)$, so $t = \frac{\ln(54)}{0.43}$.

Step5: Find the year

The starting year is 1855. The year $Y=1855 + t$. Substitute $t=\frac{\ln(54)}{0.43}\approx\frac{3.9889}{0.43}\approx9.28$. So $Y = 1855+9.28=1864.28$.

Answer:

1864.28