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.

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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**.

---
**Problem 1: Car Depreciation** - **Question:** A car depreciates at 15% per year. After how many years will it be worth half of its original value? - **Answer:** It will take approximately 4.26 years for the car to depreciate to half of its original value. **Problem 2: Bacterial Growth** - **Question:** A bacterial culture starts with 1000 bacteria and triples after 5 hours. Find the doubling period and the amount of bacteria after 8 hours. - **Answers:** - **a)** The doubling period is approximately 3.16 hours. - **b)** After 8 hours, there will be approximately 5,800 bacteria.

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.

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