e). A bacterial culture starts with 1000 bacteria and triples after 5 hours. Find the doubling period and the amount of bacteria after 8 hours.

To solve the problem, we'll model the bacterial growth using exponential functions. ### Given: - **Initial number of bacteria (N₀):** 1,000 - **Triples every 5 hours:** This means after 5 hours, the number of bacteria becomes \( 3 \times 1,000 = 3,000 \). ### 1. **Finding the Doubling Period** The doubling period (\( t_d \)) is the time it takes for the number of bacteria to double. **Step 1:** Establish the exponential growth model: \[ N(t) = N₀ \times e^{kt} \] where: - \( N(t) \) is the number of bacteria at time \( t \), - \( k \) is the growth rate constant. **Step 2:** Using the information that the bacteria triples in 5 hours: \[ 3N₀ = N₀ \times e^{5k} \] \[ 3 = e^{5k} \] \[ \ln(3) = 5k \] \[ k = \frac{\ln(3)}{5} \] **Step 3:** The doubling time (\( t_d \)) is given by: \[ t_d = \frac{\ln(2)}{k} \] \[ t_d = \frac{\ln(2)}{\ln(3)/5} \] \[ t_d = 5 \times \frac{\ln(2)}{\ln(3)} \] **Calculating Numerically:** \[ \ln(2) \approx 0.6931 \] \[ \ln(3) \approx 1.0986 \] \[ t_d \approx 5 \times \frac{0.6931}{1.0986} \approx 3.15 \text{ hours} \] ### 2. **Calculating the Number of Bacteria After 8 Hours** **Using the exponential growth model:** \[ N(t) = 1,000 \times 3^{t/5} \] **For \( t = 8 \) hours:** \[ N(8) = 1,000 \times 3^{8/5} \] \[ N(8) \approx 1,000 \times 5.805 \] \[ N(8) \approx 5,805 \text{ bacteria} \] ### **Final Answer:** - **Doubling Period:** Approximately **3.15 hours** - **Number of Bacteria After 8 Hours:** Approximately **5,805 bacteria**