To model the situation where the senior class population decreases by 5% each year, we can use an exponential decay formula. The formula for the population after \( n \) years is given by:
\[ P(n) = P_0 \times (1 - r)^n \]
where:
- \( P(n) \) is the population after \( n \) years,
- \( P_0 \) is the initial population,
- \( r \) is the rate of decrease (in decimal form),
- \( n \) is the number of years.
For this problem:
- The initial population \( P_0 = 160 \)
- The rate \( r = 0.05 \) (which represents 5%)
Thus, the equation becomes:
\[ P(n) = 160 \times (0.95)^n \]
We want to find \( n \) when \( P(n) = 100 \). So we can set up the equation:
\[ 160 \times (0.95)^n = 100 \]
To solve for \( n \), we first divide both sides by 160:
\[ (0.95)^n = \frac{100}{160} \]
This simplifies to:
\[ (0.95)^n = 0.625 \]
Next, we take the natural logarithm of both sides:
\[ \ln((0.95)^n) = \ln(0.625) \]
Using the property of logarithms, we can bring down the exponent:
\[ n \cdot \ln(0.95) = \ln(0.625) \]
Now, to solve for \( n \):
\[ n = \frac{\ln(0.625)}{\ln(0.95)} \]
Now we can calculate \( n \):
-
Calculate \( \ln(0.625) \):
\[ \ln(0.625) \approx -0.470 \]
-
Calculate \( \ln(0.95) \):
\[ \ln(0.95) \approx -0.0513 \]
-
Substitute these values into the equation for \( n \):
\[ n \approx \frac{-0.470}{-0.0513} \approx 9.15 \]
Thus, rounding to the nearest hundredth, we find:
In \( n \approx 9.15 \) years, the senior class will have about 100 students.
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