https://file.aiquickdraw.com/mathbot/file/1722092224346qso6met5.PNG

Let's solve the problems step by step.

### a) Find the exponential decay model \( P(t) = Ae^{-rt} \)

Given:
- At \( t = 5 \) minutes, \( P(5) = 38 \) micrograms per liter
- At \( t = 100 \) minutes, \( P(100) = 7 \) micrograms per liter

The general form of the exponential decay model is:
\[ P(t) = Ae^{-rt} \]

Using the given data points, we can set up two equations to solve for \( A \) and \( r \).

1. \( P(5) = 38 \)
\[ 38 = Ae^{-5r} \quad \text{(1)} \]

2. \( P(100) = 7 \)
\[ 7 = Ae^{-100r} \quad \text{(2)} \]

To solve for \( A \) and \( r \), divide equation (2) by equation (1):
\[ \frac{7}{38} = \frac{Ae^{-100r}}{Ae^{-5r}} \]
\[ \frac{7}{38} = e^{-100r} \cdot e^{5r} \]
\[ \frac{7}{38} = e^{-95r} \]

Taking the natural logarithm of both sides:
\[ \ln\left(\frac{7}{38}\right) = -95r \]
\[ -95r = \ln\left(\frac{7}{38}\right) \]
\[ r = -\frac{1}{95} \ln\left(\frac{7}{38}\right) \]

Calculate \( r \):
\[ r \approx -\frac{1}{95} \ln\left(\frac{7}{38}\right) \approx 0.0262 \]

Now substitute \( r \) back into equation (1) to find \( A \):
\[ 38 = Ae^{-5 \times 0.0262} \]
\[ 38 = Ae^{-0.131} \]
\[ A = \frac{38}{e^{-0.131}} \]
\[ A \approx 38 \cdot e^{0.131} \approx 38 \cdot 1.14 \approx 43.32 \]

Thus, the exponential decay model is:
\[ P(t) = 43.32 e^{-0.0262t} \]

### b) From the model, determine the initial amount of THC after use.

The initial amount of THC after use is \( A \):
\[ A \approx 43.32 \text{ micrograms per liter} \]

### c) Determine the half-life of THC, based on the model.

The half-life \( t_{1/2} \) is the time it takes for the concentration to reduce to half of its initial value:
\[ P(t_{1/2}) = \frac{A}{2} \]
\[ \frac{A}{2} = A e^{-rt_{1/2}} \]
\[ \frac{1}{2} = e^{-rt_{1/2}} \]
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{1}{2}\right) = -rt_{1/2} \]
\[ t_{1/2} = -\frac{\ln\left(\frac{1}{2}\right)}{r} \]
\[ t_{1/2} = \frac{\ln(2)}{r} \]
\[ t_{1/2} = \frac{\ln(2)}{0.0262} \approx 26.47 \text{ minutes} \]

### d) Determine how long after use it takes for the THC level to drop below 5 micrograms per liter.

We need to find \( t \) such that \( P(t) < 5 \):
\[ 5 = 43.32 e^{-0.0262t} \]
\[ \frac{5}{43.32} = e^{-0.0262t} \]
\[ \ln\left(\frac{5}{43.32}\right) = -0.0262t \]
\[ t = -\frac{\ln\left(\frac{5}{43.32}\right)}{0.0262} \]

Calculate \( t \):
\[ t \approx -\frac{\ln\left(\frac{5}{43.32}\right)}{0.0262} \approx 117.5 \text{ minutes} \]

### Summary of Results:
a) Exponential decay model: \( P(t) = 43.32 e^{-0.0262t} \)

b) Initial amount of THC after use: \( 43.32 \) micrograms per liter

c) Half-life of THC: \( 26.47 \) minutes

d) Time for THC level to drop below 5 micrograms per liter: \( 117.5 \) minutes

Would you like any more details or have any questions? Here are five more questions you can ask:

1. How do we derive the natural logarithm in solving exponential equations?
2. Can you explain the concept of half-life in more detail?
3. How would the model change if the decay rate were different?
4. What is the significance of the constant \( e \) in exponential models?
5. How can we apply this model to other substances or scenarios?

**Tip:** When solving exponential decay problems, carefully track your use of natural logarithms to ensure accurate calculations.

Learn how to solve the exponential decay model for THC concentration with a step-by-step guide. Calculate the initial amount of THC, determine the half-life, and find the time for THC levels to drop below 5 micrograms per liter. Suitable for college-level students and researchers.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation