To determine how long it will take for the population of a town to grow from 10,000 to 15,300 when it grows at a rate of 2.5% annually, we can use the formula for exponential growth.
The exponential growth formula is:
[tex]\[ P(t) = P_0 \times (1 + r)^t \][/tex]
Where:
- [tex]\( P(t) \)[/tex] is the future population
- [tex]\( P_0 \)[/tex] is the initial population
- [tex]\( r \)[/tex] is the growth rate (expressed as a decimal)
- [tex]\( t \)[/tex] is the time in years
Given:
- Initial population ([tex]\( P_0 \)[/tex]): 10,000
- Growth rate ([tex]\( r \)[/tex]): 2.5% or 0.025 (as a decimal)
- Target population ([tex]\( P(t) \)[/tex]): 15,300
We need to solve for [tex]\( t \)[/tex], the time in years. We start by rearranging the formula to isolate [tex]\( t \)[/tex]:
[tex]\[ P(t) = P_0 \times (1 + r)^t \][/tex]
[tex]\[ 15300 = 10000 \times (1.025)^t \][/tex]
We divide both sides by 10,000 to simplify:
[tex]\[ \frac{15300}{10000} = (1.025)^t \][/tex]
[tex]\[ 1.53 = (1.025)^t \][/tex]
To solve for [tex]\( t \)[/tex], we take the natural logarithm of both sides:
[tex]\[ \ln(1.53) = \ln((1.025)^t) \][/tex]
Using the logarithm power rule:
[tex]\[ \ln(1.53) = t \times \ln(1.025) \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(1.53)}{\ln(1.025)} \][/tex]
Using the values:
[tex]\[ \ln(1.53) \approx 0.42596873227228116 \][/tex]
[tex]\[ \ln(1.025) \approx 0.024692612590371637 \][/tex]
[tex]\[ t = \frac{0.42596873227228116}{0.024692612590371637} \approx 17.222468211815386 \][/tex]
Thus, it will take approximately 17.22 years for the population to grow to 15,300. To the nearest year, this is 17 years.
So, the town's population will reach 15,300 in approximately 17 years.
