Answer :
Certainly! Let's estimate the population of a country in 2010 using the given data and the exponential growth formula:
The exponential growth formula is given by:
[tex]\[ P(t) = P_0 e^{k t} \][/tex]
Where:
- [tex]\( P(t) \)[/tex] is the population at time [tex]\( t \)[/tex].
- [tex]\( P_0 \)[/tex] is the initial population.
- [tex]\( k \)[/tex] is the growth rate constant.
- [tex]\( t \)[/tex] is the time period.
Step-by-Step Solution:
1. Determine the initial values and given data:
- Initial population in 1992 ([tex]\( P_0 \)[/tex]) = 169 million.
- Population in 1999 ([tex]\( P(7) \)[/tex]) = 174 million. Here, [tex]\( t_1 = 1999 - 1992 = 7 \)[/tex] years.
2. Calculate the growth rate constant ([tex]\( k \)[/tex]):
Using the formula [tex]\( P(t) = P_0 e^{k t} \)[/tex], we need to determine [tex]\( k \)[/tex].
Given:
[tex]\[ P(7) = 174 \][/tex]
[tex]\[ 174 = 169 e^{k \cdot 7} \][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[ \frac{174}{169} = e^{7k} \][/tex]
Take the natural logarithm on both sides:
[tex]\[ \ln \left( \frac{174}{169} \right) = 7k \][/tex]
[tex]\[ k = \frac{\ln \left( \frac{174}{169} \right)}{7} \][/tex]
After calculating, we find:
[tex]\[ k \approx 0.004165226327350803 \][/tex]
3. Estimate the population in 2010:
Now, we want to find the population in 2010 ([tex]\( P(18) \)[/tex]). Here, [tex]\( t_2 = 2010 - 1992 = 18 \)[/tex] years.
Using the exponential growth formula again:
[tex]\[ P(18) = P_0 e^{k \cdot 18} \][/tex]
Substitute in the values:
[tex]\[ P(18) = 169 e^{0.004165226327350803 \cdot 18} \][/tex]
After calculating, we get:
[tex]\[ P(18) \approx 182.1576987981676 \][/tex]
4. Round to the nearest million:
The estimated population in 2010, rounded to the nearest million, is:
[tex]\[ P(18) \approx 182 \text{ million} \][/tex]
So, the estimated population in 2010 is 182 million.
The exponential growth formula is given by:
[tex]\[ P(t) = P_0 e^{k t} \][/tex]
Where:
- [tex]\( P(t) \)[/tex] is the population at time [tex]\( t \)[/tex].
- [tex]\( P_0 \)[/tex] is the initial population.
- [tex]\( k \)[/tex] is the growth rate constant.
- [tex]\( t \)[/tex] is the time period.
Step-by-Step Solution:
1. Determine the initial values and given data:
- Initial population in 1992 ([tex]\( P_0 \)[/tex]) = 169 million.
- Population in 1999 ([tex]\( P(7) \)[/tex]) = 174 million. Here, [tex]\( t_1 = 1999 - 1992 = 7 \)[/tex] years.
2. Calculate the growth rate constant ([tex]\( k \)[/tex]):
Using the formula [tex]\( P(t) = P_0 e^{k t} \)[/tex], we need to determine [tex]\( k \)[/tex].
Given:
[tex]\[ P(7) = 174 \][/tex]
[tex]\[ 174 = 169 e^{k \cdot 7} \][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[ \frac{174}{169} = e^{7k} \][/tex]
Take the natural logarithm on both sides:
[tex]\[ \ln \left( \frac{174}{169} \right) = 7k \][/tex]
[tex]\[ k = \frac{\ln \left( \frac{174}{169} \right)}{7} \][/tex]
After calculating, we find:
[tex]\[ k \approx 0.004165226327350803 \][/tex]
3. Estimate the population in 2010:
Now, we want to find the population in 2010 ([tex]\( P(18) \)[/tex]). Here, [tex]\( t_2 = 2010 - 1992 = 18 \)[/tex] years.
Using the exponential growth formula again:
[tex]\[ P(18) = P_0 e^{k \cdot 18} \][/tex]
Substitute in the values:
[tex]\[ P(18) = 169 e^{0.004165226327350803 \cdot 18} \][/tex]
After calculating, we get:
[tex]\[ P(18) \approx 182.1576987981676 \][/tex]
4. Round to the nearest million:
The estimated population in 2010, rounded to the nearest million, is:
[tex]\[ P(18) \approx 182 \text{ million} \][/tex]
So, the estimated population in 2010 is 182 million.