Exponential Growth and Decay
Exponential growth and decay describe amounts that change by the same percent each time period. For growth, use y = a(1 + r)^t. For decay, use y = a(1 - r)^t. Here a is the starting amount, r is the rate as a decimal, and t is the number of time periods.
Before this lesson: You should know exponential functions, how to work with percentages, and percent change.
Growth and Decay by a Percent
Say you have $100 and it grows 10% each year. After one year you have $110. The next year, 10% of $110 is $11, so you have $121. The amount added gets bigger each year.
Growing 10% is the same as multiplying by 1.10. So after t years you have 100(1.10)^t. Multiplying by the same number each year is what makes this exponential.
Decay works the same way. Losing 20% means you keep 80%. So you multiply by 0.80 each year.
Exponential Growth and Decay Formulas
\text{Growth: } y = a(1 + r)^t \qquad \text{Decay: } y = a(1 - r)^t| Symbol | Meaning | Example: 2,000 people, growing 10% per year |
|---|---|---|
| a | Starting amount (when t = 0) | a = 2000 |
| r | Rate of growth or decay, written as a decimal | r = 0.10 |
| 1 + r or 1 - r | Growth or decay factor: the number you multiply by each period | 1 + 0.10 = 1.10 |
| t | Number of time periods | Years |
The growth factor is bigger than 1. The decay factor is between 0 and 1. The rate is the percent change each period.
How to Solve a Growth or Decay Problem
- Find the starting amount, a.
- Change the percent to a decimal to get r. For example, 6% is 0.06.
- Decide if it is growth or decay. Growth: factor is 1 + r. Decay: factor is 1 - r.
- Write y = a(\text{factor})^t. Check that t counts the right time periods.
- Put in the value of t. Do the exponent first, then multiply. Include units.
Worked Examples
Example 1: Growth or Decay?
For each function, say if it is growth or decay and give the rate. (a) y = 500(1.04)^t, (b) y = 80(0.75)^t, (c) y = 12(1.5)^t.
(a) The factor is 1.04, which is more than 1. So it is growth. 1.04 = 1 + 0.04, so the rate is 4%.
(b) The factor is 0.75, which is less than 1. So it is decay. 0.75 = 1 - 0.25, so the rate is 25%.
(c) The factor is 1.5. So it is growth. 1.5 = 1 + 0.5, so the rate is 50%.
Answer: (a) growth, 4%; (b) decay, 25%; (c) growth, 50%
Example 2: Population Growth
A town has 2,000 people. Its population grows 10% each year. Write a function for the population after t years. Then find the population after 3 years.
Step 1: a = 2000 and r = 0.10. It is growth, so the factor is 1 + 0.10 = 1.1.
Step 2: P(t) = 2000(1.1)^t.
Step 3: (1.1)^3 = 1.1 \cdot 1.1 \cdot 1.1 = 1.331. Then P(3) = 2000 \cdot 1.331 = 2662.
Answer: P(t) = 2000(1.1)^t. After 3 years, the town has 2,662 people.
Example 3: Depreciation of a Car
A car costs $20,000. It loses 20% of its value each year. Write a function for its value. Then find the value after 2 years and after 3 years.
Step 1: a = 20{,}000 and r = 0.20. It is decay, so the factor is 1 - 0.20 = 0.8.
Step 2: V(t) = 20{,}000(0.8)^t.
Step 3: V(2) = 20{,}000(0.64) = 12{,}800. Then V(3) = 20{,}000(0.512) = 10{,}240.
Look at the graph. The car loses $4,000 the first year but only $3,200 the second year. That is because 20% is taken from a smaller amount each time.
Answer: After 2 years the car is worth $12,800. After 3 years it is worth $10,240.
Example 4: Half-Life
A patient takes 80 mg of a medicine. Every 6 hours, half of it leaves the body. How much is left after 24 hours?
Step 1: Half is left, so the factor is \tfrac{1}{2}. But the factor applies every 6 hours, not every hour. In t hours there are \tfrac{t}{6} half-lives.
Step 2: A(t) = 80\left(\tfrac{1}{2}\right)^{t/6}.
Step 3: For t = 24, the exponent is 24 \div 6 = 4. So A = 80\left(\tfrac{1}{2}\right)^4 = 80 \cdot \tfrac{1}{16} = 5.
Answer: 5 mg is left after 24 hours.
Example 5: Find the Rate from Data
A mobile game had 1,200 players at the start. One year later it had 900 players. Two years later it had 675. Find the rate of decay, write a function, and predict the number of players after 3 years.
Step 1: Find the factor by dividing: 900 \div 1200 = 0.75 and 675 \div 900 = 0.75. The factor is the same, so this is exponential decay.
Step 2: 0.75 = 1 - 0.25, so the rate is 25% per year. The function is N(t) = 1200(0.75)^t.
Step 3: N(3) = 675 \cdot 0.75 = 506.25. You can’t have part of a player, so round.
Answer: The game loses 25% of its players each year. After 3 years it will have about 506 players.
Common Mistakes
| Mistake | Example of the error | Fix |
|---|---|---|
| Using r as the factor | Growth of 4%: y = 500(0.04)^t | Add 1: y = 500(1.04)^t |
| Adding for decay | Decay of 20%: y = a(1.2)^t | Subtract: y = a(0.8)^t |
| Not changing the percent to a decimal | Growth of 6%: y = a(1 + 6)^t | 6% is 0.06: y = a(1.06)^t |
| Subtracting the same amount each year | 20% of 20,000 is 4,000, so year 2 is 20{,}000 - 8000 = 12{,}000 | Take 20% of the new amount: 20{,}000(0.8)^2 = 12{,}800 |
| Wrong number of periods | Half-life of 6 hours, after 24 hours: 80\left(\tfrac{1}{2}\right)^{24} | Count half-lives: 24 \div 6 = 4, so 80\left(\tfrac{1}{2}\right)^4 |
Growth Factors and Rates at a Glance
| Situation | Rate | Factor | Equation (start 100) |
|---|---|---|---|
| Grows 5% per year | 5% | 1.05 | y = 100(1.05)^t |
| Doubles each period | 100% | 2 | y = 100(2)^t |
| Shrinks 30% per year | 30% | 0.70 | y = 100(0.7)^t |
| Halves each period | 50% | 0.5 | y = 100(0.5)^t |
Money in a savings account grows the same way. When interest is added more than once a year, use the compound interest formula in the next lesson. A one-time percent increase or decrease is percent change.
Practice Problems
- Is y = 300(1.07)^t growth or decay? What is the rate?
- Is y = 45(0.9)^t growth or decay? What is the rate?
- A city has 1,500 trees. The number grows 6% each year. Write a function for the trees after t years.
- A pond has 60 fish. The number drops 15% each year. Write a function for the fish after t years.
- Find the value of 5000(1.2)^t when t = 2.
- A $900 laptop loses 10% of its value each year. What is it worth after 2 years?
- A sample has 150 bacteria. The number doubles every 3 hours. How many are there after 12 hours?
- A substance has a half-life of 5 days. You start with 64 grams. How much is left after 15 days?
- An account has $250. One year later it has $300. What is the yearly growth rate?
- A town of 8,000 people shrinks 5% each year. What is the population after 2 years?
Show answers
- Growth, because 1.07 \gt 1. The rate is 7%.
- Decay, because 0.9 \lt 1. 0.9 = 1 - 0.1, so the rate is 10%.
- T(t) = 1500(1.06)^t
- The factor is 1 - 0.15 = 0.85. So F(t) = 60(0.85)^t.
- 5000(1.2)^2 = 5000 \cdot 1.44 = 7200
- 900(0.9)^2 = 900 \cdot 0.81 = 729. It is worth $729.
- 12 hours is 12 \div 3 = 4 doubling periods. 150 \cdot 2^4 = 150 \cdot 16 = 2400 bacteria.
- 15 days is 15 \div 5 = 3 half-lives. 64\left(\tfrac{1}{2}\right)^3 = 8 grams.
- The factor is 300 \div 250 = 1.2, so the rate is 20% per year.
- 8000(0.95)^2 = 8000 \cdot 0.9025 = 7220 people.
FAQ
What is the difference between the growth rate and the growth factor?
The rate is the percent change, like 8%. The factor is what you multiply by, which is 1 + r. A rate of 8% gives a factor of 1.08.
How can you tell growth from decay in an equation?
Look at the number being raised to the power. If it is more than 1, it is growth. If it is between 0 and 1, it is decay.
What is half-life?
Half-life is the time it takes for half of an amount to be gone. After each half-life, multiply by \tfrac{1}{2}. So the exponent is the time divided by the half-life.
Why does exponential decay never reach zero?
Each period, you keep a fraction of what is left. A fraction of a positive amount is still positive. The amount gets very small but never reaches 0.
Summary
- Growth: y = a(1 + r)^t. Decay: y = a(1 - r)^t.
- a is the starting amount, r is the rate as a decimal, and t is the number of periods.
- The growth factor is more than 1. The decay factor is between 0 and 1.
- For doubling or half-life, divide the time by the length of one period to get the exponent.
- Each period takes the percent of the new amount, not the starting amount.
Previous lesson: Exponential Functions
Next lesson: Compound Interest
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