To find out how many years it took for the number of students to reach 365, we can set up the equation using the function given:
\[ f(n) = 240 (1.15)^n \]
We need to find \( n \) such that:
\[ 240 (1.15)^n = 365 \]
First, we can divide both sides by 240:
\[ (1.15)^n = \frac{365}{240} \]
Calculating the right side:
\[ \frac{365}{240} \approx 1.52083333 \]
Now, we can take the logarithm of both sides to solve for \( n \):
\[ \log((1.15)^n) = \log(1.52083333) \]
Using the power rule of logarithms:
\[ n \log(1.15) = \log(1.52083333) \]
Now solve for \( n \):
\[ n = \frac{\log(1.52083333)}{\log(1.15)} \]
Calculating the logarithms:
- \( \log(1.52083333) \approx 0.1818 \)
- \( \log(1.15) \approx 0.0672 \)
Now substitute these values into the equation for \( n \):
\[ n \approx \frac{0.1818}{0.0672} \approx 2.70 \]
Since \( n \) must be a whole number and we need to find when the number of students reaches at least 365, we can check for \( n = 2 \) and \( n = 3 \):
For \( n = 2 \):
\[ f(2) = 240 (1.15)^2 \approx 240 (1.3225) \approx 317.4 \]
For \( n = 3 \):
\[ f(3) = 240 (1.15)^3 \approx 240 (1.5209) \approx 364.97 \]
Since \( f(3) \approx 364.97 \) is just below 365 and f(4) would exceed 365, we find that the first year when the student numbers reached or exceeded 365 is after the third year.
Thus, the answer is:
C. 3 years.
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