Let's solve the problem step-by-step.
We are given:
- The initial enrollment, [tex]\( P_0 = 12000 \)[/tex] students.
- The final enrollment, [tex]\( P_t = 15000 \)[/tex] students.
- The continuous growth rate, [tex]\( r = 2.5\% = 0.025 \)[/tex] per year.
We are asked to determine the logarithm expression that equates to the number of years, [tex]\( t \)[/tex], it takes for the population to grow from 12,000 to 15,000 students.
The formula for continuous growth is given by:
[tex]\[ P_t = P_0 e^{rt} \][/tex]
Here, [tex]\( t \)[/tex] is the number of years we need to find.
1. First, substitute the known values into the formula:
[tex]\[ 15000 = 12000 e^{0.025t} \][/tex]
2. Next, solve for [tex]\( e^{0.025t} \)[/tex]:
[tex]\[ \frac{15000}{12000} = e^{0.025t} \][/tex]
3. Simplify the fraction:
[tex]\[ \frac{15000}{12000} = 1.25 \][/tex]
[tex]\[ 1.25 = e^{0.025t} \][/tex]
4. Take the natural logarithm ([tex]\( \ln \)[/tex]) of both sides to linearize the equation:
[tex]\[ \ln(1.25) = \ln(e^{0.025t}) \][/tex]
5. Use the property of logarithms that [tex]\( \ln(e^x) = x \)[/tex]:
[tex]\[ \ln(1.25) = 0.025t \][/tex]
6. Finally, solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(1.25)}{0.025} \][/tex]
Hence, the correct logarithm expression that represents the number of years it will take for the population to increase to 15,000 students is:
[tex]\[ \frac{\ln 1.25}{0.025} \][/tex]
So the correct choice is:
[tex]\[ \boxed{4} \][/tex]
