Question

SA model for the number \(N\) of people in a college community who have heard a certain rumor is$$N(d)=P\left(1-e^{-0.15 d}\right)$$ where \(P\) is the total population of the community and \(d\) is the number of days that have elapsed since the rumor began. In a community of 1000 students, how many days will elapse before 450 students have heard the rumor?

Short answer

It will take approximately 4.3 days for 450 students to hear the rumor.

Step-by-step solution

- Understand the Given Formula

The formula given is \[ N(d) = P \big( 1 - e^{-0.15 d} \big) \]. Here, \( N(d) \) is the number of people who have heard the rumor after \( d \) days, and \( P \) is the total population of the community.

- Plug in the Known Values

We know \( P = 1000 \) and we need to find \( d \) such that \( N(d) = 450 \). Substitute these into the equation: \[ 450 = 1000 \big( 1 - e^{-0.15 d} \big) \]

- Rearrange to Isolate the Exponential Term

To isolate the exponential term, divide both sides of the equation by 1000: \[ \frac{450}{1000} = 1 - e^{-0.15 d} \] Simplify the fraction: \[ 0.45 = 1 - e^{-0.15 d} \] Then rearrange to solve for the exponential term: \[ e^{-0.15 d} = 1 - 0.45 \] \[ e^{-0.15 d} = 0.55 \]

- Solve for \( d \) Using Natural Logarithms

Take the natural logarithm of both sides to solve for \( d \): \[ \text{ln} \big( e^{-0.15 d} \big) = \text{ln}(0.55) \] Simplify using the properties of logarithms: \[ -0.15 d = \text{ln}(0.55) \]

- Final Calculation

Divide both sides by -0.15 to isolate \( d \): \[ d = \frac{\text{ln}(0.55)}{-0.15} \] Calculate the value using a calculator: \[ d \frac{\text{ln}(0.55)}{-0.15} \ \rightarrow d \ \rightarrow d \ \rightarrow Calculate Using a Calculator \]

Key concepts

rumor spreading model

In this problem, we are working with a rumor spreading model representing how information spreads in a community. The model is given by the equation: \( N(d) = P ( 1 - e^{-0.15 d} ) \) where \( N(d) \) denotes the number of people who have heard the rumor after \( d \) days, and \( P \) stands for the total population of the community. This model follows the principle of exponential growth and decay.

The rumor spreads quickly at the beginning when many people are unaware, but the rate slows down as more people hear about it. In other words, the number of people hearing the rumor increases rapidly at first and then levels off as it reaches the entire population. This kind of behavior is typical for many natural phenomena, such as population growth, radioactive decay, and in this case, spreading information.

natural logarithms

Natural logarithms, denoted as \( \text{ln}(x) \), are logarithms with the base \( e \), where \( e \approx 2.71828 \). They are incredibly useful in solving exponential equations. In this exercise, we use natural logarithms to solve for the number of days it takes for 450 students to hear the rumor. Here's a brief overview of how we applied the natural logarithm:

After rearranging the equation \( 450 = 1000 (1 - e^{-0.15 d}) \) to isolate the exponential term \( e^{-0.15 d} = 0.55 \), we take the natural logarithm of both sides:

\( \text{ln}(e^{-0.15 d}) = \text{ln}(0.55) \)

Using the property of logarithms \( \text{ln}(e^x) = x \), this simplifies to:

\( -0.15 d = \text{ln}(0.55) \)

Natural logarithms simplify the process of solving exponential equations by converting the multiplicative processes into additive ones, thereby allowing us to isolate the variable easily.

solving exponential equations

Solving exponential equations involves steps to isolate the exponential variable, often using logarithms. Let's review the process:

1. **Isolate the exponential term:** Start by isolating the term with the exponent. Here, we rearranged \( 450 = 1000 (1 - e^{-0.15 d}) \) to get \( e^{-0.15 d} = 0.55 \).

2. **Apply the natural logarithm:** Take the natural logarithm of both sides to remove the exponent from the variable: \( \text{ln}(e^{-0.15 d}) = \text{ln}(0.55) \).

3. **Simplify the logarithms:** Using properties of logarithms, simplify to isolate the variable: \( -0.15 d = \text{ln}(0.55) \).

4. **Solve for the variable:** Divide by the coefficient to find \( d \): \( d = \frac{\text{ln}(0.55)}{-0.15} \).

Using this process, we discovered that it takes approximately 5.3 days for 450 students to hear the rumor. This step-by-step approach is fundamental for solving many similar problems in mathematics, making it a crucial skill in your mathematical toolkit.