Certainly! Let's tackle this problem step-by-step.
### Given:
- The population in 2003 is 54,000.
- The rate of increase is 5% per annum.
We can use the formula for population growth:
[tex]\[ P(t) = P_0 \times (1 + r)^t \][/tex]
where:
- [tex]\( P(t) \)[/tex] is the population at time [tex]\( t \)[/tex].
- [tex]\( P_0 \)[/tex] is the initial population.
- [tex]\( r \)[/tex] is the growth rate (5% or 0.05).
- [tex]\( t \)[/tex] is the number of years from the initial measurement.
### 1. Finding the population in 2001:
Since 2001 is 2 years before 2003, we have [tex]\( t = -2 \)[/tex].
The formula becomes:
[tex]\[ 54000 = P_0 \times (1 + 0.05)^{-2} \][/tex]
We need to solve for [tex]\( P_0 \)[/tex]:
[tex]\[ P_0 = \frac{54000}{(1 + 0.05)^{-2}} \][/tex]
Evaluating this:
[tex]\[ P_0 = \frac{54000}{(1.05)^{-2}} \][/tex]
[tex]\[ P_0 = \frac{54000}{1 / (1.05)^2} \][/tex]
[tex]\[ P_0 = 54000 \times (1.05)^2 \][/tex]
Calculating the value, we get:
[tex]\[ P_0 \approx 48979.59183673469 \][/tex]
So, the population in 2001 was approximately 48,979.59.
### 2. Finding the population in 2005:
Since 2005 is 2 years after 2003, we have [tex]\( t = 2 \)[/tex].
The formula becomes:
[tex]\[ P(2005) = 54000 \times (1 + 0.05)^2 \][/tex]
Evaluating this:
[tex]\[ P(2005) = 54000 \times (1.05)^2 \][/tex]
Calculating the value, we get:
[tex]\[ P(2005) \approx 59535.0 \][/tex]
So, the population in 2005 would be approximately 59,535.
### Summary:
1. The population in 2001 was approximately 48,979.59.
2. The population in 2005 would be approximately 59,535.
