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.