How to Write an Exponential Function: Word Problems

To write an exponential function from a word problem, find two things: the starting amount a and the factor b that multiplies the quantity each period. For growth of r percent, b = 1 + r; for decay of r percent, b = 1 − r. Then write y = a·bˣ. For example, $500 growing 4% a year is y = 500(1.04)ˣ.

Exponential functions model anything that grows or shrinks by the same percent or factor each period: populations, savings, loans, car values, bacteria and radioactive decay. This guide shows how to turn the words into y = a·b^x, with eight worked examples covering growth, decay, compound interest, doubling, half-life and finding the factor from data.

The exponential function

y = a · b^x

• a = the starting (initial) amount, the value when x = 0
• b = the growth or decay factor, what the amount is multiplied by each period
• x = the number of periods (years, hours, months)

For a percent change of r (written as a decimal):

How to write the function: 4 steps

1. Find a: the amount at the start.
2. Find b: is it growing or shrinking, and by what percent or factor each period?
3. Define x: what one period is (one year, one hour).
4. Write y = a·b^x, then substitute to answer the question and check that the answer makes sense.

Worked examples

1. Growth. A bank’s profits start at $60,000 and increase by 25% each year. How much after 4 years?
a = 60,000 and b = 1.25, so y = 60,000(1.25)^x. After 4 years: y = 60,000(1.25)⁴ = 60,000 × 2.4414 ≈ $146,484.38.

2. Population. A town of 5,000 grows 3% per year. Write the function and find the population after 10 years.
y = 5,000(1.03)^x; after 10 years y ≈ 6,720 people.

3. Depreciation (decay). A car bought for $24,000 loses 15% of its value each year.
b = 1 − 0.15 = 0.85, so y = 24,000(0.85)^x. After 3 years: 24,000 × 0.614125 ≈ $14,739.

4. Savings. $800 earns 4% interest compounded yearly. y = 800(1.04)^x; after 5 years ≈ $973.32.

5. Compound interest more than once a year. $1,000 at 6% compounded monthly for 5 years: A = P(1 + r/n)^nt = 1,000(1 + 0.06/12)⁶⁰ ≈ $1,348.85, a little more than yearly compounding ($1,338.23).

6. Doubling. A culture of 200 bacteria doubles every hour. y = 200(2)^x; after 6 hours, y = 200 × 64 = 12,800.

7. Half-life. 80 grams of a substance has a half-life of 5 years. y = 80(1/2)^t/5; after 15 years (three half-lives) y = 80 × 1/8 = 10 grams.

8. Finding b from data. A population is 45 in year 2 and 405 in year 4. Over 2 years it multiplied by 405 ÷ 45 = 9, so each year it multiplies by √9 = 3: b = 3. Working back one year at a time, the year-0 value is 45 ÷ 3 ÷ 3 = 5, so y = 5(3)^x.

Exponential or linear? Read the wording

Common mistakes

• Using the percent as the factor: 25% growth is b = 1.25, not 25 or 0.25.
• Using 1 + r for decay: a 15% loss is 0.85, not 1.15.
• Mixing time units: if the rate is per year and the question is in months, convert first.
• Rounding b too early. Keep the exact factor until the last step.

Practice problems

1. A city of 12,000 grows 2% per year. Write the function and estimate the population after 5 years.
2. A phone worth $900 loses 20% of its value each year. What is it worth after 2 years?
3. A sample of 64 mg has a half-life of 3 days. How much is left after 9 days?
4. Is “a salary rises $1,500 each year” linear or exponential?

Answers

1. y = 12,000(1.02)^x; after 5 years ≈ 13,249.
2. 900(0.8)² = $576.
3. Three half-lives: 64 × 1/8 = 8 mg.
4. Linear (same amount each year).

Frequently Asked Questions

How do you write an exponential function from a word problem?

Find the starting amount a and the factor b that multiplies the quantity each period, then write y = a·b^x. For growth of r%, b = 1 + r (as a decimal); for decay of r%, b = 1 − r.

How do you know if a word problem is exponential or linear?

If the quantity changes by the same amount each period (adds $50 a month), it is linear. If it changes by the same percent or factor each period (grows 5% a year, doubles every hour), it is exponential.

What is the growth factor for 8% growth?

1.08. Write the percent as a decimal (0.08) and add it to 1. For 8% decay the factor is 1 − 0.08 = 0.92.

How do you write a half-life or doubling-time function?

Use y = a·(1/2)^t/h for a half-life of h, or y = a·2^t/d for a doubling time of d. The exponent t/h counts how many half-lives have passed.

Related: Modeling with functions · Comparing linear functions · Euler’s method · Domain and range of quadratics