A car depreciates at \( 15 \% \) per year. After how many years will it be worth half of its original value? 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.

Let's tackle both problems step by step. --- ## Problem 1: Car Depreciation **Statement:** A car depreciates at **15%** per year. After how many years will it be worth half of its original value? ### **Solution:** Depreciation can be modeled using an exponential decay formula: \[ V(t) = V_0 \times (1 - r)^t \] where: - \( V(t) \) = value of the car after \( t \) years - \( V_0 \) = original value of the car - \( r \) = annual depreciation rate (in decimal) - \( t \) = time in years Given: - \( r = 15\% = 0.15 \) - We want \( V(t) = \frac{1}{2} V_0 \) **Setting up the equation:** \[ \frac{1}{2} V_0 = V_0 \times (0.85)^t \] Divide both sides by \( V_0 \): \[ \frac{1}{2} = 0.85^t \] **Solving for \( t \):** Take the natural logarithm on both sides to solve for \( t \): \[ \ln\left(\frac{1}{2}\right) = t \times \ln(0.85) \] \[ t = \frac{\ln(0.5)}{\ln(0.85)} \] **Calculating the value:** \[ t \approx \frac{-0.6931}{-0.1625} \approx 4.26 \text{ years} \] ### **Answer:** It will take approximately **4.26 years** for the car to depreciate to half of its original value. --- ## Problem 2: Bacterial Growth **Statement:** A bacterial culture starts with **1000 bacteria** and triples after **5 hours**. a) Find the **doubling period**. b) Find the **amount of bacteria after 8 hours**. ### **Solution:** This is a problem of exponential growth, which can be modeled by: \[ N(t) = N_0 \times e^{kt} \] where: - \( N(t) \) = number of bacteria at time \( t \) - \( N_0 \) = initial number of bacteria - \( k \) = growth constant - \( t \) = time in hours **Given:** - \( N_0 = 1000 \) bacteria - After 5 hours, \( N(5) = 3 \times 1000 = 3000 \) bacteria ### **a) Finding the Doubling Period** We need to find the time \( T \) it takes for the bacteria to double, i.e., \( N(T) = 2N_0 \). **Setting up the equation:** \[ 2N_0 = N_0 \times e^{kT} \] \[ 2 = e^{kT} \] Take the natural logarithm on both sides: \[ \ln(2) = kT \] \[ T = \frac{\ln(2)}{k} \] **First, find \( k \):** From the tripling information: \[ 3000 = 1000 \times e^{5k} \] \[ 3 = e^{5k} \] Take natural logarithm: \[ \ln(3) = 5k \] \[ k = \frac{\ln(3)}{5} \approx \frac{1.0986}{5} \approx 0.2197 \text{ per hour} \] **Now, calculate the doubling period \( T \):** \[ T = \frac{\ln(2)}{0.2197} \approx \frac{0.6931}{0.2197} \approx 3.16 \text{ hours} \] ### **b) Amount of Bacteria After 8 Hours** Using the growth formula: \[ N(8) = 1000 \times e^{k \times 8} \] Substitute \( k \): \[ N(8) = 1000 \times e^{0.2197 \times 8} = 1000 \times e^{1.7576} \] Calculate \( e^{1.7576} \): \[ e^{1.7576} \approx 5.80 \] Thus: \[ N(8) \approx 1000 \times 5.80 = 5800 \text{ bacteria} \] ### **Answers:** - **a)** The **doubling period** is approximately **3.16 hours**. - **b)** After **8 hours**, there will be approximately **5,800 bacteria**. ---