This mini-lesson covers AQA section F — Exponentials and logarithms: the graphs of ax and ex, why e is special, logarithms and their laws, ln as the inverse of ex, solving equations of the form ax = b, growth and decay models, and the crucial technique of reducing non-linear data to y = mx + c using logs.
Work through each screen, answer the questions as you go (a few are conceptual, most are calculations) and collect ⭐ stars. Everything here is A-level standard, drawn from the AQA 7357 subject content. Press Start when you are ready.
An exponential function has the variable in the index: y = ax with a > 0. Every such curve passes through (0, 1) and has the x-axis (y = 0) as a horizontal asymptote — it never reaches zero.
That property is why e is the natural base for calculus. More generally, by the chain rule:
Growth vs decay: in y = Aekt, a positive k means growth and a negative k means decay. In y = 800e−0.2t, k = −0.2, so the quantity decays.
A logarithm answers the question "what power?":
So log₂ 32 = 5, log₁₀ 1000 = 3, and log₁₀ 0.01 = −2 (because 10⁻² = 0.01).
ln x means loge x — the natural logarithm. It is the inverse of ex, so:
You cannot take the log of a negative number or of zero. The domain of ln x is x > 0. Any solution that requires ln(negative) must be rejected.
Three laws, all inherited directly from the laws of indices:
Also worth knowing: loga 1 = 0 and loga a = 1.
2 log₃ 6 − log₃ 4 = log₃ 36 − log₃ 4 = log₃ (36/4) = log₃ 9 = 2.
And log₃ 9 + log₃ 3 = 2 + 1 = 3 (or, in one step, log₃ 27 = 3).
The two fatal errors: log(a + b) is not log a + log b, and log a / log b is not log(a/b). The laws only apply to log of a product, quotient or power.
Tap a model, then tap how it behaves as t increases.
When the unknown is in the index, take logs of both sides. Any base works; ln is usually cleanest.
Take ln: ln(3x) = ln 20 ⇒ x ln 3 = ln 20 (using k log a = log aᵏ).
x = ln 20 / ln 3 = 2.9957… / 1.0986… = 2.727 (3 d.p.).
Check: 32.727 = 19.998 ≈ 20 ✓
Exponentiate both sides: 2x − 1 = e³ = 20.0855…
2x = 21.0855… ⇒ x = 10.54 (2 d.p.). And 2x − 1 > 0 ✓, so the solution is valid.
Hidden quadratics: e2x − 5ex + 6 = 0 is a quadratic in y = ex: y² − 5y + 6 = 0 ⇒ y = 2 or 3 ⇒ x = ln 2 or ln 3. If a root of the quadratic is negative, reject it — ex is never negative.
Whenever a rate of change is proportional to the amount present, the model is exponential:
A sample decays as N = 200 e−0.05t (t in years). When does it fall to 50?
50 = 200 e−0.05t ⇒ e−0.05t = 0.25
−0.05t = ln 0.25 = −1.3863 ⇒ t = 1.3863 / 0.05 = 27.7 years (1 d.p.).
Sense check: halving takes ln2/0.05 ≈ 13.9 years, and 50 is two halvings from 200 — about 27.7 years ✓
Interpret the constants in context. N₀ is the starting value; k is the growth or decay rate; and a decay model has a horizontal asymptote at N = 0 — the model never quite reaches zero, which is a genuine limitation worth stating.
This is the technique AQA loves in the large data set and in modelling questions. Two model types, two different plots.
The distinction is the whole point:
A straight line on a log–log plot is evidence for a power law. That is what such graphs are actually telling you — and it is exactly what the question means by "explain how the graph supports the model".
Once you have the straight line, you must undo the logs to get a and b back.
Data believed to fit y = a xb. Plotting log₁₀y against log₁₀x gives a straight line with gradient 1.5 and vertical intercept 0.7.
Comparing with log y = b log x + log a:
b = gradient = 1.5.
log₁₀ a = 0.7 ⇒ a = 100.7 = 5.01 (3 s.f.).
So the model is y = 5.01 x1.5.
The intercept is log a, not a. Forgetting to raise 10 (or e) to the power of the intercept is the standard error here — and it is always worth a mark.
Tap an item on the left, then its partner on the right.
Know the shapes cold, since sketch questions are common.
Transformed asymptotes move. For y = 5 − 2e−x, as x → ∞ the exponential term vanishes, so the asymptote is y = 5 — and the curve approaches it from below.
Exponentials: y = aˣ passes through (0, 1), asymptote y = 0 · d/dx(e^{kx}) = k e^{kx}
Logs: log_a b = c ⇔ aᶜ = b · ln is log base e · ln(eˣ) = x
Log laws: log a + log b = log(ab) · log a − log b = log(a/b) · k log a = log aᵏ
Equations: take logs when x is in the index; watch for hidden quadratics in eˣ
Models: N = N₀e^{kt} — k > 0 growth, k < 0 decay
Linearising: y = axᵇ → plot log y vs log x · y = abˣ → plot log y vs x — and the intercept is log a, not a
That is AQA section F — Exponentials and logarithms. Press Finish to see your score.
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