Answer :
To estimate the population in 2005 using the exponential growth formula, we start with the given data:
- Population in 1991, [tex]\( P_{1991} \)[/tex] = 15 million
- Population in 2001, [tex]\( P_{2001} \)[/tex] = 16 million
- Time elapsed between 1991 and 2001, [tex]\( t_1 \)[/tex] = 2001 - 1991 = 10 years
The exponential growth formula is [tex]\( P = A e^{k t} \)[/tex], where:
- [tex]\( P \)[/tex] is the population at time [tex]\( t \)[/tex]
- [tex]\( A \)[/tex] is the initial population
- [tex]\( k \)[/tex] is the growth rate
- [tex]\( t \)[/tex] is the time
We have two known points:
1. At [tex]\( t_1 = 0 \)[/tex], [tex]\( P = P_{1991} \)[/tex] = 15 million
2. At [tex]\( t_1 = 10 \)[/tex] (from 1991 to 2001), [tex]\( P = P_{2001} \)[/tex] = 16 million
Using these points, we can solve for the growth rate [tex]\( k \)[/tex].
First, apply the exponential growth formula to the second point:
[tex]\[ P_{2001} = P_{1991} \cdot e^{k \cdot t_1} \][/tex]
[tex]\[ 16 = 15 \cdot e^{k \cdot 10} \][/tex]
Now, solve for [tex]\( k \)[/tex]:
[tex]\[ \frac{16}{15} = e^{10k} \][/tex]
[tex]\[ \ln\left(\frac{16}{15}\right) = 10k \][/tex]
[tex]\[ k = \frac{\ln\left(\frac{16}{15}\right)}{10} \][/tex]
Calculating the natural logarithm and dividing by 10 gives:
[tex]\[ k \approx \frac{\ln(1.0667)}{10} \][/tex]
[tex]\[ k \approx \frac{0.0645}{10} \][/tex]
[tex]\[ k \approx 0.0065 \][/tex] (rounded to four decimal places)
Next, estimate the population for 2005. The time elapsed from 1991 to 2005 is [tex]\( t_2 = 2005 - 1991 = 14 \)[/tex] years.
Now, use the exponential growth formula for [tex]\( t_2 \)[/tex]:
[tex]\[ P_{2005} = P_{1991} \cdot e^{k \cdot t_2} \][/tex]
[tex]\[ P_{2005} = 15 \cdot e^{0.0065 \cdot 14} \][/tex]
Now, calculate [tex]\( P_{2005} \)[/tex]:
[tex]\[ P_{2005} \approx 15 \cdot e^{0.091} \][/tex]
[tex]\[ P_{2005} \approx 15 \cdot 1.0954 \][/tex]
[tex]\[ P_{2005} \approx 16.4184 \][/tex]
Finally, round the estimated population to the nearest million:
[tex]\[ P_{2005} \approx 16 \][/tex]
Therefore, the estimated population in 2005 is 16 million.
- Population in 1991, [tex]\( P_{1991} \)[/tex] = 15 million
- Population in 2001, [tex]\( P_{2001} \)[/tex] = 16 million
- Time elapsed between 1991 and 2001, [tex]\( t_1 \)[/tex] = 2001 - 1991 = 10 years
The exponential growth formula is [tex]\( P = A e^{k t} \)[/tex], where:
- [tex]\( P \)[/tex] is the population at time [tex]\( t \)[/tex]
- [tex]\( A \)[/tex] is the initial population
- [tex]\( k \)[/tex] is the growth rate
- [tex]\( t \)[/tex] is the time
We have two known points:
1. At [tex]\( t_1 = 0 \)[/tex], [tex]\( P = P_{1991} \)[/tex] = 15 million
2. At [tex]\( t_1 = 10 \)[/tex] (from 1991 to 2001), [tex]\( P = P_{2001} \)[/tex] = 16 million
Using these points, we can solve for the growth rate [tex]\( k \)[/tex].
First, apply the exponential growth formula to the second point:
[tex]\[ P_{2001} = P_{1991} \cdot e^{k \cdot t_1} \][/tex]
[tex]\[ 16 = 15 \cdot e^{k \cdot 10} \][/tex]
Now, solve for [tex]\( k \)[/tex]:
[tex]\[ \frac{16}{15} = e^{10k} \][/tex]
[tex]\[ \ln\left(\frac{16}{15}\right) = 10k \][/tex]
[tex]\[ k = \frac{\ln\left(\frac{16}{15}\right)}{10} \][/tex]
Calculating the natural logarithm and dividing by 10 gives:
[tex]\[ k \approx \frac{\ln(1.0667)}{10} \][/tex]
[tex]\[ k \approx \frac{0.0645}{10} \][/tex]
[tex]\[ k \approx 0.0065 \][/tex] (rounded to four decimal places)
Next, estimate the population for 2005. The time elapsed from 1991 to 2005 is [tex]\( t_2 = 2005 - 1991 = 14 \)[/tex] years.
Now, use the exponential growth formula for [tex]\( t_2 \)[/tex]:
[tex]\[ P_{2005} = P_{1991} \cdot e^{k \cdot t_2} \][/tex]
[tex]\[ P_{2005} = 15 \cdot e^{0.0065 \cdot 14} \][/tex]
Now, calculate [tex]\( P_{2005} \)[/tex]:
[tex]\[ P_{2005} \approx 15 \cdot e^{0.091} \][/tex]
[tex]\[ P_{2005} \approx 15 \cdot 1.0954 \][/tex]
[tex]\[ P_{2005} \approx 16.4184 \][/tex]
Finally, round the estimated population to the nearest million:
[tex]\[ P_{2005} \approx 16 \][/tex]
Therefore, the estimated population in 2005 is 16 million.
