Question

an initial census taken in a small asian country found the population to be 23.9 million. ten years later, the population had increased to 24.7 million. during the same time period, a census was taken in a south pacific nation. the initial census found a population of 20.0 million. ten years later, the population had increased to 23.6 million.
(a) write a function of the form ( p ( t ) = p _ { 0 } e ^ { k t } ) to model the population ( p ( t ) ) (in millions) ( t ) years after the initial census for the asian and south pacific countries. round the value of ( k ) to five decimal places.

Explanation:

Step1: Identify the initial population \(P_0\)

For the Asian country, when \(t = 0\), \(P(0)=P_0 = 23.9\) (from the table). So the population function is \(P(t)=23.9e^{kt}\).

Step2: Use the data at \(t = 10\) to find \(k\)

When \(t = 10\), \(P(10)=24.7\). Substitute into \(P(t)=23.9e^{kt}\):
\(24.7=23.9e^{10k}\).
First, divide both sides by \(23.9\): \(\frac{24.7}{23.9}=e^{10k}\).
Then, take the natural - logarithm of both sides: \(\ln(\frac{24.7}{23.9})=\ln(e^{10k})\).
Since \(\ln(e^{x}) = x\), we have \(10k=\ln(\frac{24.7}{23.9})\).
Calculate \(\ln(\frac{24.7}{23.9})\approx\ln(1.03347)\approx0.033\).
So \(k=\frac{\ln(\frac{24.7}{23.9})}{10}\approx\frac{0.033}{10}=0.0033\).

Answer:

\(P(t)=23.9e^{0.0033t}\)