How to Solve Exponential Growth and Decay Functions?
Exponential growth and decay describe situations where a quantity increases or decreases by a constant percentage each time period. Unlike linear change (which adds the same amount each period), exponential change multiplies by the same factor, leading to rapid growth or rapid decline. Population growth, radioactive decay, compound interest, and the spread of viruses are all classic examples.
Solve Exponential Growth and Decay Functions: what to notice and how to work it
What to notice first
Common student mistake
Key formulas and cues
A reliable path
- Find the startLook for the value when x or time is 0.
- Find the multiplierUse the base or percent change to identify b.
- Check the shapeGrowth rises away from the asymptote; decay moves toward it.
Worked examples
Growth model
- Start at 100.
- Multiply by 2 three times.
- Compute 100 times 8.
Decay model
- Half of 80 is 40.
- Half of 40 is 20.
- This is multiplying by 1/2 twice.
Try one before moving on
Solve Exponential Growth and Decay Functions: pop-up practice
What Are Exponential Growth and Decay?
In exponential growth, a quantity increases by a fixed percent each period — the amount added each time gets larger and larger. In exponential decay, a quantity decreases by a fixed percent each period — the amount removed gets smaller and smaller over time.
The Exponential Growth and Decay Formula
\(\color{blue}{A = P(1 \pm r)}\)t
Where:
- A = Final amount after time t
- P = Initial (starting) amount (principal)
- r = Rate of change per period (as a decimal; divide percent by 100)
- t = Number of time periods
- Use + r for growth; use − r for decay
Identifying Growth vs. Decay
Exponential Growth: \(\color{blue}{A = P(1 + r)}\)t
The base \(\color{blue}{(1 + r)}\) is greater than 1. The value of A increases as t increases.
- P = $1,000, \(\color{blue}{r = 5}\)% = 0.05, \(\color{blue}{t = 3}\): \(\color{blue}{A = 1000(1.05)^{3} \approx $1,157.63}\).
Exponential Decay: \(\color{blue}{A = P(1 – r)}\)t
The base \(\color{blue}{(1 – r)}\) is between 0 and 1. The value of A decreases as t increases.
- P = $2,000, \(\color{blue}{r = 10}\)% = 0.10, \(\color{blue}{t = 4}\): \(\color{blue}{A = 2000(0.90)^{4} \approx $1,312.20}\).
Step-by-Step Summary
- Identify whether the quantity is growing (\(\color{blue}{\text{ use } +r}\)) or decaying (\(\color{blue}{\text{ use } -r}\)).
- Convert the rate to a decimal.
- Substitute P, r, and t into the formula \(\color{blue}{A = P(1 \pm r)^{t}}\).
- Use a calculator or exponent rules to evaluate \(\color{blue}{(1 \pm r)^{t}}\).
- Multiply by P to find the final amount A.
Watch: Graphing Exponential Growth & Decay
Khan Academy explains the shape and behavior of exponential growth and decay graphs:
Exponential Growth and Decay – Worked Examples
Example 1 (Growth): A savings account has $1,000 at 5% annual growth. What is the balance after 3 years?
\(\color{blue}{A = 1000(1 + 0.05)^{3} = 1000(1.05)^{3} = 1000 \times 1.157625 \approx $1,157.63}\).
Example 2 (Decay): A car purchased for $2,000 loses 10% of its value each year. What is its value after 4 years?
\(\color{blue}{A = 2000(1 – 0.10)^{4} = 2000(0.90)^{4} = 2000 \times 0.6561 \approx $1,312.20}\).
Example 3 (Growth): A town of 500 people grows at 8% per year. What is the population after 5 years?
\(\color{blue}{A = 500(1.08)^{5} \approx 500 \times 1.46933 \approx 734.66}\) (approximately 735 people).
Example 4 (Decay): A radioactive sample of 3,000 grams decays at 15% per year. How much remains after 3 years?
\(\color{blue}{A = 3000(1 – 0.15)^{3} = 3000(0.85)^{3} = 3000 \times 0.614125 \approx $1,842.37 g}\).
More Practice: Exponential Growth and Decay Word Problems
Khan Academy solves a variety of exponential growth and decay word problems:
Exercises for Exponential Growth and Decay
Use \(\color{blue}{A = P(1 \pm r)}\)t to find the final amount. Round to the nearest cent.
- Growth: P = $1,500, \(\color{blue}{r = 6}\)%, \(\color{blue}{t = 4}\) years.
- Decay: P = $3,000, \(\color{blue}{r = 15}\)%, \(\color{blue}{t = 3}\) years.
- Growth: P = $800, \(\color{blue}{r = 10}\)%, \(\color{blue}{t = 5}\) years.
- Decay: P = $5,000, \(\color{blue}{r = 20}\)%, \(\color{blue}{t = 2}\) years.
- Growth: P = $250, \(\color{blue}{r = 4}\)%, \(\color{blue}{t = 10}\) years.
Answers
- \(\color{blue}{A \approx $1,893.72}\)
- \(\color{blue}{A \approx $1,842.37}\)
- \(\color{blue}{A \approx $1,288.41}\)
- \(\color{blue}{A = $3,200.00}\)
- \(\color{blue}{A \approx $370.06}\)
Free Exponential Growth and Decay Worksheet
Ready to practice on your own? Download our free Exponential Growth and Decay worksheet below, work through each problem at your own pace, and then check your answers. If a few give you trouble, scroll back up to the worked examples and try again — steady practice is the surest way to master Exponential Growth and Decay before a quiz or test.
Download Exponential Growth Worksheet
Frequently Asked Questions
What is the difference between simple interest and exponential growth?
Simple interest adds the same dollar amount each period (linear). Exponential growth multiplies by the same factor each period, so the dollar amount added grows over time. Compound interest is an example of exponential growth.
How do I tell if a function shows growth or decay from its equation?
In \(\color{blue}{A = P \times b^{t}}\), if \(\color{blue}{b > 1}\) it is growth; if \(\color{blue}{0 < b < 1}\) it is decay. In the \(\color{blue}{A = P(1 \pm r)^{t}}\) form, a “+” means growth and a “−” means decay.
Can I use this formula for population problems?
Yes. Population growth at a constant rate follows the exponential growth model. Replace P with the initial population and t with the number of years (or other time periods).
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