Exponential growth and decay follow the model y = abˣ, where a is the starting value and b is the multiplier per period. When b > 1, y grows; when 0 < b < 1, y decays. Unlike linear change, each period multiplies the value by the same factor — not adds the same amount.
What is the difference between exponential and linear change?
In linear growth, the quantity increases by the same amount each period. In exponential growth, it increases by the same percentage — meaning the absolute increase gets larger every period.
| Period | Linear growth (+10 each time) | Exponential growth (×1.1 each time) |
|---|---|---|
| 0 | 100 | 100 |
| 1 | 110 | 110 |
| 2 | 120 | 121 |
| 3 | 130 | 133.1 |
| 5 | 150 | 161.1 |
| 10 | 200 | 259.4 |
After just 10 periods, exponential growth is nearly 30% higher than linear growth, and the gap widens rapidly.
What does the formula y = abˣ mean?
- a is the initial value (the value of y when x = 0, since b⁰ = 1).
- b is the growth or decay factor per period.
- x is the number of periods elapsed.
| Situation | b value | Example |
|---|---|---|
| Growth of 5% per year | b = 1.05 | Savings with 5% interest |
| Decay of 10% per year | b = 0.90 | Car value decreasing 10%/yr |
| Doubles every period | b = 2 | Bacteria doubling every hour |
| Halves every period | b = 0.5 | Radioactive half-life |
How do you use the model to solve problems?
Worked example 1 — bacteria growth:
A colony starts with 500 bacteria. The number doubles every hour. How many bacteria are there after 4 hours?
y = abˣ = 500 × 2⁴ = 500 × 16 = 8000 bacteria
Worked example 2 — radioactive decay:
A radioactive substance has a mass of 200 g. Its mass halves every 5 years (half-life = 5 years). Find the mass after 20 years.
After 20 years, there are 20 ÷ 5 = 4 half-lives.
y = 200 × (0.5)⁴ = 200 × 0.0625 = 12.5 g
Worked example 3 — finding a and b from two data points:
A population was 4000 in 2010 and 4802 in 2012. Find the annual growth factor b, then estimate the population in 2015.
Let x = years after 2010.
y = ab^x.
When x = 0: a = 4000.
When x = 2: 4802 = 4000 × b².
b² = 4802/4000 = 1.2005 → b = √1.2005 ≈ 1.096
Check: 4000 × 1.096 ≈ 4384 (2011); 4384 × 1.096 ≈ 4805 (2012) ≈ 4802 ✓ (rounding)
2015 is x = 5: y = 4000 × 1.096⁵ ≈ 4000 × 1.585 ≈ 6340 people
How do you sketch an exponential graph?
The graph of y = abˣ (for b > 1) has these features:
- It passes through (0, a) — the y-intercept is a.
- It is always above the x-axis (never touches or crosses it).
- It curves steeply upward to the right.
- The x-axis is an asymptote: as x → −∞, y → 0 but never reaches 0.
For decay (0 < b < 1), the same bullet points hold except the curve falls steeply to the right.
Key points to plot: (0, a), (1, ab), (2, ab²). Connect smoothly with a curve that does not touch the x-axis.
What is the connection to compound interest?
Compound interest is exponential growth in disguise. The compound interest formula A = P(1 + r/100)ⁿ is exactly y = abˣ with a = P, b = (1 + r/100) and x = n. Similarly, compound depreciation uses b = (1 − r/100). The key conceptual link: both compound interest and exponential growth multiply by the same factor each period.
Frequently asked questions
How is exponential decay different from compound depreciation?
They are the same mathematical model — y = abˣ with 0 < b < 1. In the context of asset values, the b factor is called a depreciation multiplier (e.g. 0.85 for 15% annual depreciation). In science, the same calculation describes radioactive decay or drug concentration in the blood. The numbers and context differ; the equation is identical.
Can I use logarithms to solve exponential equations at GCSE?
No. Logarithms are not part of the GCSE specification (they appear at A-level). At GCSE, exponential equations are solved by trial and improvement, reading off a graph, or by substituting convenient values when the answer is a whole number of periods (e.g., the number of doublings). You will not be asked to solve 2ˣ = 7 exactly at GCSE.
What does it mean when a question says "how many periods for the quantity to halve"?
This is asking for the half-life. Substitute y = a/2 into y = abˣ: a/2 = abˣ → b^x = 0.5. Trial-and-improve x by calculating powers of b until you reach 0.5. For b = 0.9: 0.9¹ = 0.9, 0.9⁷ ≈ 0.478, 0.9⁶ ≈ 0.531 — so the half-life is between 6 and 7 periods. Read off the graph for a non-integer answer.
How do I recognise an exponential function from a table of values?
Compute the ratio y₂/y₁, y₃/y₂, y₄/y₃ for consecutive rows. If all ratios are equal (constant multiplier), the function is exponential. If the differences y₂−y₁, y₃−y₂ are equal, it is linear. This is a standard exam technique.
Explore exponential growth step by step with Professor Pi at aitutors.me.