Note:
When solving for [tex]$k$[/tex], round to four decimal places.

A country's population in 1995 was 56 million. In 2002, it was 59 million. Estimate the population in 2016 using the exponential growth formula. Round your answer to the nearest million.

[tex] P = A e^{k t} [/tex]

Enter the correct answer:
[tex]\boxed{\phantom{enter\ the\ correct\ answer}}[/tex]



Answer :

To estimate the population of a country in 2016 given its population in 1995 and 2002, we will use the exponential growth formula:

[tex]\[ P = A e^{k t} \][/tex]

Here, \( P \) is the population at time \( t \), \( A \) is the initial population, \( e \) is the base of the natural logarithm, \( k \) is the growth rate, and \( t \) is the time elapsed.

1. Initial Values:
- Population in 1995: \( P_{1995} = 56 \) million
- Population in 2002: \( P_{2002} = 59 \) million
- Time interval from 1995 to 2002: \( t_1 = 2002 - 1995 = 7 \) years

2. Calculate the growth rate \( k \):
- Using the formula for growth rate in terms of populations and time:

[tex]\[ P_{2002} = P_{1995} e^{k \cdot t_1} \][/tex]

Solving for \( k \):

[tex]\[ k = \frac{\ln\left(\frac{P_{2002}}{P_{1995}}\right)}{t_1} = \frac{\ln\left(\frac{59}{56}\right)}{7} \][/tex]

Evaluating the natural logarithm and dividing it by the time interval:

[tex]\[ k \approx \frac{\ln(1.0536)}{7} \approx \frac{0.0522}{7} \approx 0.0075 \][/tex]

Thus, \( k \approx 0.0075 \) (rounded to four decimal places).

3. Estimate the population in 2016:
- Time interval from 1995 to 2016: \( t_2 = 2016 - 1995 = 21 \) years
- Using the formula:

[tex]\[ P_{2016} = P_{1995} e^{k \cdot t_2} \][/tex]

Substituting the values:

[tex]\[ P_{2016} = 56 e^{0.0075 \cdot 21} \][/tex]

Evaluating the exponent:

[tex]\[ P_{2016} \approx 56 e^{0.1575} \approx 56 \cdot 1.1716 \approx 65.4907525510204 \][/tex]

Rounding to the nearest million:

[tex]\[ P_{2016} \approx 65 \text{ million} \][/tex]

So, the estimated population of the country in 2016 is:

[tex]\[ \boxed{65} \][/tex]