To determine how many years it will take for the population of the town to grow from 17,000 to 23,200 with an annual growth rate of 3%, we can use the formula for exponential 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 (as a decimal),
- [tex]\( t \)[/tex] is the time in years.
We are given:
- Initial population [tex]\( P_0 = 17000 \)[/tex],
- Final population [tex]\( P(t) = 23200 \)[/tex],
- Growth rate [tex]\( r = 3\% = 0.03 \)[/tex].
We can rearrange the formula to solve for [tex]\( t \)[/tex]:
[tex]\[ 23200 = 17000 \times (1 + 0.03)^t \][/tex]
Dividing both sides by 17000 to isolate the exponential term:
[tex]\[ \frac{23200}{17000} = (1 + 0.03)^t \][/tex]
[tex]\[ \frac{23200}{17000} = 1.03^t \][/tex]
Next, we take the natural logarithm (ln) of both sides to solve for [tex]\( t \)[/tex]:
[tex]\[ \ln\left(\frac{23200}{17000}\right) = \ln(1.03^t) \][/tex]
Using the property of logarithms that states [tex]\( \ln(a^b) = b \cdot \ln(a) \)[/tex]:
[tex]\[ \ln\left(\frac{23200}{17000}\right) = t \cdot \ln(1.03) \][/tex]
Now, we can solve for [tex]\( t \)[/tex] by dividing by [tex]\( \ln(1.03) \)[/tex]:
[tex]\[ t = \frac{\ln\left(\frac{23200}{17000}\right)}{\ln(1.03)} \][/tex]
[tex]\[ t = \frac{\ln\left(\frac{23200}{17000}\right)}{\ln(1.03)} \][/tex]
Calculating this expression using a calculator or software that can compute natural logarithms:
[tex]\[ t \approx \frac{\ln(1.36470588235)}{\ln(1.03)} \][/tex]
[tex]\[ t \approx \frac{0.3101549284}{0.0295588022} \][/tex]
[tex]\[ t \approx 10.488 \][/tex]
After calculating, the result is approximately 10.488 years. Since we need to find the number of years to the nearest year, we round 10.488 to the nearest whole number, which is 10.
Therefore, it will take approximately 10 years for the population to reach 23,200, to the nearest year.
