Exponential growth and decay: word problems
key notes :
🎯 Understanding Exponential Growth and Decay
- Exponential growth happens when a quantity increases by a constant percentage over equal time periods.
- Exponential decay happens when a quantity decreases by a constant percentage over equal time periods.
- The general formula for both is:
A = A0 × (1±r)t
Where:
- A = final amount
- A0 = initial amount
- r = rate of increase or decrease (as a decimal)
- t = time (usually in years, but can be other units)
- + for growth, – for decay
🔍Key Characteristics
✅ Exponential Growth:
- The value increases rapidly after a certain point (e.g., population growth, investments).
- Graph rises steeply.
- Example: Bacteria doubling every hour.
✅ Exponential Decay:
- The value decreases but never quite hits zero (e.g., radioactive decay, depreciation).
- Graph falls but flattens over time.
- Example: Half-life of a substance.
🛠️Steps to Solve Word Problems
1️⃣ Identify the type — Is it growth (+) or decay (-)?
2️⃣ Identify the variables — What are the initial amount, rate, and time?
3️⃣ Set up the formula — Plug the values into the formula correctly.
4️⃣ Solve carefully — Be cautious with powers and decimals.
5️⃣ Check the answer — Does it make sense? (e.g., a population can’t be negative!)
🔥Example Word Problems
✅ Example 1: Population Growth
A town’s population is 5,000 and grows at a rate of 2% per year. What will the population be in 10 years?
A = 5000 × (1+0.02)10
A = 5000 × (1+0.02)10
A ≈ 5000 × 1.219 = 6095
👉 Answer: The population will be about 6,095 people.
✅ Example 2: Radioactive Decay
A radioactive substance has a mass of 120 grams and decays at a rate of 5% per hour. How much is left after 8 hours?
A = 120 × (1−0.05)8
A = 120 × (0.95)8
A ≈ 120×0.663 = 79.56 grams
👉 Answer: About 79.56 grams remain.
✅ Example 3: Investment Growth
A $1,000 investment grows at 6% annually. How much will it be worth after 5 years?
A = 1000 × (1+0.06)5
A = 1000 × (1.06)5
A ≈ 1000 × 1.338=1338
👉 Answer: The investment will grow to $1,338.
Learn with an example
🗼 A city’s population is currently 150,000. If the population doubles every 30 years, what will the population be 60 years from now?
- First, find out how many times the population will double. Divide the number of years by how long it takes for the population to double.
- 60 ÷ 30 = 2
- The population will double 2 times.
- Now figure out what the population will be after it doubles 2 times. Multiply the population by 2 a total of 2 times.
- 150,000 × 2 × 2 = 600,000
- That calculation could also be written with exponents:
- 150,000 × 22 = 600,000
- After 60 years, the population will be 600,000 people.
🗼 A virus is spreading around Westminster. At the moment, 490 people are infected. If the virus is spreading at a rate of 15% each day, how many people in all will have caught the virus in 14 days?
If necessary, round your answer to the nearest whole number.
_________people
- Plug in 490 people for the initial amount, 0.15 for the rate of increase, and 14 days for the time elapsed.
| y | = | a(1 + r)t | |
| y | = | 490(1 + 0.15)14 | Plug in |
| y | = | 490(1.15)14 | Add |
| y | ≈ | 490(7.075) | Simplify |
| y | ≈ | 3,466.75 | Multiply |
- Round to the nearest whole number.
| 3,466.75 | → | 3,467 |
- To the nearest whole number, 3,467 people will have caught the virus.
🗼 Sam has put ₹1,800,000 into a pension. The fund has an estimated annual return of 5%. If Sam doesn’t add any more money, how much can he expect to have in the fund in 20 years?
If necessary, round your answer to the nearest paisa.
₹________
- Plug in ₹1,800,000 for the initial amount, 0.05 for the rate of increase, and 20 years for the time elapsed.
| y | = | a(1 + r)t | |
| y | = | 1,800,000(1 + 0.05)20 | Plug in |
| y | = | 1,800,000(1.05)20 | Add |
| y | ≈ | 1,800,000(2.653297705) | Simplify |
| y | ≈ | 4,775,935.869 | Multiply |
- Round to the nearest paisa.
| 4,775,935.869 | → | 4,775,935.87 |
- To the nearest paisa, Sam can expect to have ₹4,775,935.87 in his pension.
Let’s practice!🖊️
