Linearise an exponential model and interpret growth or decay
Plot log y against the unlogged input, recover a positive base and distinguish a negative rate from a negative amount.
Section 3.2 · Logarithmic and exponential functions
Exam scope and assessment notes
Pure Mathematics 2 (Paper 2, AS only) and Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027. Paper 1 knowledge is assumed. The AS route using Papers 1 + 2 cannot be carried forward to A Level. Shared sections 2.2 and 3.2 include log laws excluding change of base, natural exponential and logarithmic graphs, equations and inequalities, and straight-line transformations. Written practice is self-checked.
Before you begin
- Use log laws and find a line's gradient and intercept.
- Distinguish a positive exponential base from the sign of its logarithm.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which axes make y = kaˣ a straight line using natural logs?
What you will learn
- Transform y = kaˣ into ln y = ln k + x ln a and recover k and a.
- Interpret the slope of a log-output plot as growth, decay or a constant model, and check assumptions.
For an exponential model, transform only the output axis
For y = kaˣ with k > 0 and a > 0, ln y = ln k + x ln a. Plot vertical Y = ln y against horizontal X = x; the slope is ln a and the intercept is ln k. Hence a = e^(slope) and k = e^(intercept). If the model is y = ke^(bx), its slope is b directly. The horizontal input stays unlogged, unlike the two transformed axes for a power model.
A line through (x, ln y) = (0, ln 5) and (2, ln 20) has gradient [ln 20 − ln 5]/2 = (ln 4)/2 = ln 2. The intercept gives k = 5 and the gradient gives a = 2, so y = 5·2ˣ. Both constants are recovered by undoing a logarithm. The raw points (0, 5) and (2, 20) then verify the assumed model.
The slope sign describes direction, not a negative base
A positive slope m gives a = eᵐ > 1 and exponential growth; a negative slope gives 0 < a < 1 and decay. A zero slope gives a = 1 and a constant positive model. A negative intercept also produces a positive multiplier k below 1, rather than a negative amount. In ln y = −x/4 + ln 8, the model is y = 8e^(−x/4) and the per-unit base is e^(−1/4).
Two positive observations can determine a and k under an assumed real-input exponential form. For y(1) = 6 and y(3) = 54, the ratio gives a² = 9. Keep a = 3 because the base is positive, then k = 2. This predicts y(2) = 18 between the supplied inputs. A matching pair of observations alone does not establish model validity elsewhere; retain the positive output condition and inspect further data before making a wider empirical claim.
Pause and explain
A plot of ln y against x has negative slope m. What does a = eᵐ satisfy?
Worked example
A plot of ln y against x passes through (0, ln 5) and (2, ln 20). Assuming y = kaˣ, find the model.
Show the worked solution
- The gradient is ln a = (ln 20 − ln 5)/2 = ln 2.
- Thus a = 2, and intercept ln k = ln 5 gives k = 5.
- The model y = 5·2ˣ reproduces y(0) = 5 and y(2) = 20.
Answer k = 5, a = 2; model y = 5·2ˣ.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A transformed model is ln y = (ln 3)x + ln 2. Recover y = kaˣ and find y(2).
Give me a hint
Exponentiate the slope and intercept separately.
Compare my reasoning
- The slope ln a = ln 3 gives a = 3.
- The intercept ln k = ln 2 gives k = 2, so y = 2·3ˣ.
- At x = 2 the amount is 2(9) = 18.
y = 2·3ˣ; y(2) = 18.
Look for these in your work
- Undo both logarithmic coefficients.
- Use the recovered raw model for the prediction.
A line on axes (x, ln y) passes through (0, ln 8) and (3, 0). Recover the exponential model and predict y(4).
Give me a hint
The second vertical value is ln 1 = 0; use its slope and the first point's intercept.
Compare my reasoning
- The slope is (0 − ln 8)/3 = −ln 2, giving a = 1/2.
- The intercept gives k = 8, so y = 8(1/2)ˣ, a positive decay model.
- At x = 4 it predicts y = 8/16 = 1/2; this is a model prediction beyond the two fitted input values.
y = 8(1/2)ˣ; predicted y(4) = 1/2.
Look for these in your work
- Recover a positive base below 1 from the negative slope.
- Interpret a raw prediction without claiming further empirical validation.
Positive observations y(1) = 6 and y(3) = 54 are assumed to follow y = kaˣ for real x, with positive k and a. Find k and a, predict y(2) and explain why a = −3 is excluded.
Give me a hint
Use the ratio to find a² and retain the positive-base condition before recovering k.
Compare my reasoning
- The ratio gives 54/6 = a², so the algebraic candidates are ±3.
- The model requires a > 0 for all real inputs, so use a = 3; from 6 = ka obtain k = 2.
- Then y(2) = 2(9) = 18. The value −3 is outside the assumed positive-base real-input model; further observations are needed to assess the fitted form.
a = 3, k = 2; predicted y(2) = 18; a = −3 violates the positive-base real-input model.
Look for these in your work
- Use the stated model's base domain after solving the ratio equation.
- Recover the multiplier and verify both observations.
Common mistakes
- For an exponential relation, plot log y against x, rather than automatically logging both axes.
- Undo both log coefficients to recover k and a; a negative slope still gives a positive base.
- Keep a > 0 for the real-input model and distinguish fitting from validation.
What can you explain now?
Recover an exponential model from ln y = −x/4 + ln 8 and state its direction.
Compare with the explanation
y = 8e^(−x/4), with a = e^(−1/4); it decays.
The negative slope is the natural exponential rate, while undoing the intercept gives positive multiplier 8.
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Make it stick
Compare the transformed axes for power and exponential models. Recover a positive base from a negative slope and explain the distinction between a fitted prediction and model validation.
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