Question

As a town gets smaller, the population of its high school decreases by 5% each year. The senior class has about 160 students now. In how many years will it have about 100 students? Write an equation. Then solve the equation without graphing.
Write an equation to represent this situation. Let n be the number of years before the class will have 100 students. ____ (Type an equation using n as the variable. Use integers or decimals for any numbers in the equation.
Solve the equation without graphing.
In n=____ years the senior class will have about 100 students. (Type an integer or decimal rounded to the nearest hundredth as needed.)

Answer 1

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 \( n \):

1. Calculate \( \ln(0.625) \):

\[
\ln(0.625) \approx -0.470
\]

2. Calculate \( \ln(0.95) \):

\[
\ln(0.95) \approx -0.0513
\]

3. 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.**