Assume the carrying capacity of the earth is 18 billion Use the 1960 s peak annual growth rate of $2.1 \%$ and population of 3 billion to predict the base growth rate and current growth rate with a logistic model Assume a current population of 6.8 billion. How does the predicted growh rate compare to the actual growth rate of about $1.2 \%$ per year?

What is the base growth rate?
$\square \%$ (Round to four doclmat places as needed)

Step 1 ：Given the logistic model for population growth: \(P(t) = \frac{K}{1 + Ae^{-rt}}\), where \(P(t)\) is the population at time \(t\), \(K\) is the carrying capacity, \(A\) is a constant related to the initial population, \(r\) is the base growth rate, and \(t\) is time.

Step 2 ：We know that the carrying capacity \(K\) is 18 billion, the population in 1960 was 3 billion (which we'll take as our initial population \(P_0\)), and the growth rate in 1960 was 2.1%.

Step 3 ：First, we'll find \(A\). Rearranging the logistic model equation for \(A\) gives: \(A = \frac{K - P_0}{P_0}\)

Step 4 ：Substituting the given values, we get: \(A = \frac{18 \text{ billion} - 3 \text{ billion}}{3 \text{ billion}}\)

Step 5 ：\(\boxed{A = 5}\)

Step 6 ：Next, we'll find \(r\). The growth rate in 1960 was 2.1%, which we'll write as 0.021. This is equal to \(r(1 - P_0/K)\), so we can solve for \(r\): \(0.021 = r(1 - \frac{3 \text{ billion}}{18 \text{ billion}})\)

Step 7 ：Solving the equation, we get: \(0.021 = r(1 - \frac{1}{6})\)

Step 8 ：Solving for \(r\), we get: \(r = 0.021 \times \frac{6}{5}\)

Step 9 ：\(\boxed{r = 0.0252}\), or 2.52% when expressed as a percentage.