2. Exponential Growth & Decay
What makes something "exponential"?
Here's the key distinction that trips people up: linear change adds or subtracts the same amount every step (like +£5 every week). Exponential change multiplies by the same factor every step (like ×1.05 every week — 5% more each time).
That's why exponential change feels "slow, then suddenly explosive" — each new increase is bigger than the last, because you're multiplying an ever-growing number by the same factor, not adding a fixed chunk to it.
Analogy: Imagine a rumour. On day 0, 1 person knows it. If it doubles every day (a linear idea — "add 1 person" — would be way too slow), by day 10 that's 210 = 1024 people. That explosive jump from "barely noticeable" to "everyone knows" is the signature of exponential growth.
Real-life growth examples: population increase, bacteria multiplying, viral infection spread, compound interest.
Real-life decay examples: a hot drink cooling down, a car's value depreciating, radioactive decay, medicine leaving your bloodstream.
The model: y = ax
y = ax
y = output value · a = scale factor per time period · x = number of time periods
- If a > 1, you're multiplying by something bigger than 1 each time → the value grows.
- If 0 < a < 1 (a fraction/decimal less than 1), you're multiplying by something smaller than 1 each time → the value shrinks toward zero.
- a can never be negative in these models — a negative base would make the graph flip signs every step, which doesn't make physical sense for population, mass, temperature, etc.
Worked Example — Growth
y = 3x models a fly population, y, after x days.
Day 0: 30 = 1 fly
Day 2: 32 = 9 flies
Day 4: 34 = 81 flies
Notice how the jump from day 2→4 (72 more flies) is way bigger than day 0→2 (8 more flies). Same time gap, bigger jump — that's exponential growth in action.
Worked Example — Decay
y = 0.5x models the radius (cm) of an evaporating water droplet after x minutes.
Minute 0: 0.50 = 1 cm
Minute 2: 0.52 = 0.25 cm
Minute 4: 0.54 = 0.0625 cm
The droplet shrinks fast at first, then the shrinking slows down — it keeps halving whatever is left, so it gets closer and closer to 0 but (in the maths) never quite gets there.
Practice Question 1
A large population of birds increases exponentially according to N = 1.04t, where N is measured in thousands and t is the number of years. Find the population after 13 years, to the nearest hundred.
Practice Question 2
The percentage of concert tickets still available follows y = ax, where y is the decimal proportion remaining and x is days since release. After 3 days, 51.2% remain. (a) Find a. (b) What % remains after 2 weeks (3 s.f.)?
3. Logarithms
What is a logarithm, really?
A logarithm is the inverse function of an exponential. If an exponential asks "what do I get if I raise a to the power x?", a logarithm asks the reverse question: "what power do I need to raise a to, in order to get b?"
Analogy: Think of exponentials and logs like multiplication and division, or like addition and subtraction — they're a matched pair that undo each other. If 5 × 3 = 15, then division lets you ask "15 ÷ 5 = ?" to get back to 3. Similarly, if 53 = 125, then a logarithm lets you ask "log base 5 of 125 = ?" to get back to 3.
ax = b ⟺ x = loga(b)
These two statements say exactly the same thing — just rearranged. "a" is called the base of the logarithm.
Reading it out loud: "log base a of b" is pronounced exactly like it looks — it's asking "what power do I raise a to, to get b?"
If you see log(50) with no small base number, it always means base 10. This used to be the standard base before calculators, and your calculator's plain "log" button is always base 10.
Building intuition: evaluating logs by hand
Let's build up a feel for this with base 5, working out what power gets us to different numbers:
| Expression | Reasoning | Value |
| log5(125) | 5³ = 125, so the power needed is 3 | 3 |
| log5(25) | 5² = 25 | 2 |
| log5(5) | 5¹ = 5 | 1 |
| log5(1) | 5⁰ = 1 (anything to power 0 is 1) | 0 |
| log5(1/5) | 5⁻¹ = 1/5 (negative powers → reciprocals) | −1 |
| log5(1/25) | 5⁻² = 1/25 | −2 |
| log5(√5) | 5^(1/2) = √5 (fractional powers → roots) | 1/2 |
| log5(∛5) | 5^(1/3) = ∛5 | 1/3 |
Notice the pattern: negative logs come from fractions less than 1, and fractional logs come from roots. This isn't a coincidence — it's the same index laws you already know, just viewed from the other direction.
Most calculators have a "log..." or "log_b" button that lets you type a custom base — different from the plain "log" button (which is fixed at base 10). Use it to double-check your hand calculations.
Converting between log form and exponential form
This is the single most useful skill in this whole chapter — being able to flip fluently between the two forms. Here's the trick that makes it click:
Memory trick: Read loga(b) = x as "x is the power you raise a to, to get b." Once you say that sentence in your head, converting to ax = b becomes automatic — you're just writing down the sentence in symbols.
Worked Example
The graph y = 3x passes through the point where y = 81. Find x.
Set up: 3x = 81
Convert to log form: x = log3(81)
Since 3⁴ = 81, we get x = 4
Solving exponential equations with logs
Some exponential equations solve "by inspection" — you can just see the answer. For example, 2x = 8 is obviously x = 3, because you know 2³ = 8. But what about 2x = 10? There's no whole number that works — this is where logs become essential.
Method A: ax = b ⟹ x = loga(b)
Take log base a of both sides directly (needs the "custom base" calculator button)
Method B: ax = b ⟹ x = log(b) ÷ log(a)
Using base-10 logs only (the plain "log" button) — gives the identical answer
Worked Example — Both Methods
Solve 4x = 20, correct to 3 s.f.
Method A: x = log4(20) = 2.1609... ≈ 2.16
Method B: x = log(20) ÷ log(4) = 2.1609... ≈ 2.16
Same answer either way — use whichever method your calculator supports more easily, or whichever the mark scheme in your syllabus favours.
Worked Example — No Calculator
Solve log(8x) = 4 without a calculator.
No base shown → base 10: log10(8x) = 4
Convert to exponential form: 10⁴ = 8x
Simplify: 10000 = 8x
Divide both sides by 8: x = 1250
Application: Compound interest & "how many years?"
A classic exam scenario: you know the growth rate and the target value, and you need to find how many time periods (n) it takes to get there. This is exactly the situation logs were built for.
Worked Example
You invest £500 at 5% interest per year. After how many whole years does it first exceed £900?
Set up: 500 × 1.05ⁿ = 900
Isolate the power: 1.05ⁿ = 900/500 = 1.8
Convert to logs: n = log1.05(1.8) = log(1.8)/log(1.05) = 12.047...
Since n must be a whole year and 12.047 means it hasn't quite exceeded £900 at year 12, round up to n = 13 years
In "how many years until it exceeds..." questions, always round up to the next whole year — even if the decimal is very close to the year below. 12.05 years still means year 12 hasn't reached the target yet, so you need year 13.
Application: Finding inverse functions
Since logs undo exponentials, they're exactly the tool you need whenever you're asked to find the inverse of an exponential function.
Worked Example
Find the inverse of f(x) = 24x.
Write as: y = 24x
Swap x and y: x = 24y
Convert to log form: 4y = log2(x)
Divide by 4: f⁻¹(x) = ¼ log2(x)
Practice Question 3
Make t the subject of 7p(t−1) = 2q.
Practice Question 4
Without a calculator, evaluate log2(1/8). Explain your reasoning in one line.