Let's solve each example step by step using the given exponential growth and decay formulas.
---
GROWTH FUNCTIONS
Formula:
$$
y = a(1 + r)^t
$$
Where:
- $ a $ = initial amount
- $ r $ = growth rate (as a decimal)
- $ t $ = time in years
---
####
Example 1: Population Growth
> The population of a town is currently 8,250. It has been increasing 3% each year. What will the population be in 10 years?
Step 1: Identify values
- $ a = 8250 $
- $ r = 3\% = 0.03 $
- $ t = 10 $
Step 2: Plug into formula
$$
y = 8250(1 + 0.03)^{10} = 8250(1.03)^{10}
$$
Step 3: Calculate
First compute $ (1.03)^{10} $:
$$
(1.03)^{10} \approx 1.343916
$$
Now multiply:
$$
y \approx 8250 \times 1.343916 \approx 11048.77
$$
Answer: The population after 10 years will be approximately
11,049 people.
---
####
Example 2: Salary Growth
> The starting salary at a job is $24,000. The job offers an increase of 1.5% each year. How much will an employee make after 6 years?
Step 1: Identify values
- $ a = 24000 $
- $ r = 1.5\% = 0.015 $
- $ t = 6 $
Step 2: Use formula
$$
y = 24000(1 + 0.015)^6 = 24000(1.015)^6
$$
Step 3: Compute
$$
(1.015)^6 \approx 1.093443
$$
$$
y \approx 24000 \times 1.093443 \approx 26,242.63
$$
Answer: The employee will make approximately
$26,242.63 after 6 years.
---
DECAY FUNCTIONS
Formula:
$$
y = a(1 - r)^t
$$
Where:
- $ a $ = initial amount
- $ r $ = decay rate (as a decimal)
- $ t $ = time in years
---
####
Example 1: Car Depreciation
> Yasser bought a car in 2012 for $16,000. The car’s value has been decreasing at a rate of 7% each year. How much was the car worth in 2018?
Step 1: Find time
From 2012 to 2018 is $ t = 6 $ years.
Step 2: Identify values
- $ a = 16000 $
- $ r = 7\% = 0.07 $
- $ t = 6 $
Step 3: Use formula
$$
y = 16000(1 - 0.07)^6 = 16000(0.93)^6
$$
Step 4: Compute
$$
(0.93)^6 \approx 0.6856
$$
$$
y \approx 16000 \times 0.6856 \approx 10,969.60
$$
Answer: The car was worth approximately
$10,969.60 in 2018.
---
####
Example 2: Stock Depreciation
> A stock investment is depreciating at a rate of 3% per year. If the initial investment was $12,000, how much will it be worth after 8 years?
Step 1: Identify values
- $ a = 12000 $
- $ r = 3\% = 0.03 $
- $ t = 8 $
Step 2: Use formula
$$
y = 12000(1 - 0.03)^8 = 12000(0.97)^8
$$
Step 3: Compute
$$
(0.97)^8 \approx 0.7837
$$
$$
y \approx 12000 \times 0.7837 \approx 9,404.40
$$
Answer: The investment will be worth approximately
$9,404.40 after 8 years.
---
✔ Summary of Answers:
| Example | Problem | Answer |
|--------|--------|--------|
| Growth 1 | Town population in 10 years |
11,049 |
| Growth 2 | Salary after 6 years |
$26,242.63 |
| Decay 1 | Car value in 2018 |
$10,969.60 |
| Decay 2 | Stock value after 8 years |
$9,404.40 |
These problems demonstrate real-world applications of
exponential growth and decay, commonly used in finance, biology, economics, and more.
Parent Tip: Review the logic above to help your child master the concept of practice worksheet exponential functions.
