Overview
What you’ll learn
- Reduce y = axⁿ to linear form by plotting lg y against lg x.
- Reduce y = kbˣ to linear form by plotting lg y against x.
- Choose axes to linearise other relations, such as y = a√x + b or xy = ax² + b.
- Read gradient and intercept from a line of best fit and convert them into the original constants.
- Use a linear-law graph to estimate values and verify a proposed model.
In 30 seconds
- 1A straight line has form Y = mX + c; linear law redefines X and Y until the relation takes that shape.
- 2For y = axⁿ, take logarithms: lg y = n lg x + lg a — plot lg y (vertical) against lg x (horizontal).
- 3In that plot the gradient is n and the vertical intercept is lg a.
- 4For y = kbˣ: lg y = (lg b) x + lg k — plot lg y against x; gradient lg b, intercept lg k.
- 5lg b is not b — undo the logarithm with 10^ (power of ten) at the very end.
- 6Simple swaps work without logs: y = a√x + b becomes linear in √x; y² = ax³ + b becomes linear in x³.
- 7Experimental points scatter: draw a line of best fit and read the gradient from two well-separated points on the line, not from data points.
- 8A point that sits far off the best-fit line is an outlier to investigate, not to force onto the line.
Notes · 1 of 5
Why linearise a curve?
Straight lines are the only graphs whose slope and intercept can be read by eye.
Two unknown constants hide inside y = axⁿ: the coefficient a and the index n. A curve offers no straightedge — there is no visual "slope of a parabola" to measure with a ruler. But Y = mX + c is different: its gradient m and intercept c can be read off graph paper directly. Linear law is the art of renaming the axes so the curve becomes that line.
The renaming must be exact algebra, not guesswork: choose X and Y as functions of x and y, substitute, and check that every term collects into Y = mX + c. If a stray x remains outside the chosen X, the choice is wrong. This chapter shows the two standard logarithm cases and then the general swap-in method.
Notes · 2 of 5
Power laws: plotting lg y against lg x
y = axⁿ becomes lg y = n·lg x + lg a — gradient n, intercept lg a.
Take logarithms of both sides of y = axⁿ and apply the laws: lg y = lg(axⁿ) = lg a + n lg x. Comparing with Y = mX + c: Y = lg y, X = lg x, gradient m = n, intercept c = lg a. Both unknowns fall out of one straight-line graph — n from the slope, a from 10 raised to the intercept.
Notes · 3 of 5
Exponentials: plotting lg y against x
y = kbˣ becomes lg y = (lg b)·x + lg k — gradient lg b, intercept lg k.
Now the index is the variable itself: lg(kbˣ) = lg k + x lg b. In the plot of lg y against x (not lg x), the gradient is lg b and the intercept is lg k. This is why the two cases must not be mixed up: a power law straightens with logs on both axes, an exponential with a log on the y-axis only.
With experimental scatter, draw the line of best fit rather than joining points, and compute the gradient from two points on the line far apart — using two data points instead spreads their individual errors into the answer.
Notes · 4 of 5
Beyond the two standard forms
Any relation that can be collected into Y = mX + c is fair game.
| Relation | Plot (vertical vs horizontal) | Gradient | Intercept |
|---|---|---|---|
| y = axⁿ | lg y vs lg x | n | lg a |
| y = kbˣ | lg y vs x | lg b | lg k |
| y = a√x + b | y vs √x | a | b |
| y² = ax³ + b | y² vs x³ | a | b |
| xy = ax² + b | xy vs x² | a | b |
| 1/y = ax + b | 1/y vs x | a | b |
The method is always the same: rearrange until one combination of the variables (with all constants on the other side) looks like Y = mX + c. For y = ax + b/x, the relation is stubborn as it stands, but multiplying by x gives xy = ax² + b, linear in the pair (x², xy). Sums of unlike powers, such as y = x² + 3x, refuse to linearise: no single X kills both terms.
Notes · 5 of 5
Using the recovered model
Interpolate with confidence, extrapolate with care, and check the fit before you trust it.
Once the constants are known, the model answers questions the data never asked. Interpolation — estimating between the smallest and largest measured values — is the safe case: substitute into the recovered relation. A model y = 2√x + 5 fitted on 1 ≤ x ≤ 25 handles x = 16 without any apology: y = 13.
Extrapolation beyond the data range deserves a warning label: the straight-line fit says nothing about a regime change outside the measurements. State the estimate and flag it as an extrapolation. Finally, a model earns trust only if the scatter about the line looks random; a systematic bend in the residuals means the proposed form was wrong, and no amount of best-fit drawing will fix it.
Revise
Key terms
The words examiners expect you to use precisely.
- Linear form
- A rewriting Y = mX + c where Y and X are functions of the original variables.
- Line of best fit
- The straight line drawn through scattered experimental points balancing the deviations.
- Gradient
- The slope m of the straight line in the redefined axes; usually one unknown constant (or its logarithm).
- Intercept
- The value where the line crosses the vertical axis; usually the other constant (or its logarithm).
- Outlier
- A data point far from the line of best fit, often caused by experimental error.
- Power law
- A relation of the form y = axⁿ, linearised by logs on both axes.
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- Semi-logarithmic plot
- A graph where only the vertical axis is logarithmic (lg y against x) — the natural home of exponentials.
- Interpolation
- Estimating between known data values, where the fitted line is trustworthy.
Revise
Common mistakes
Where students most often lose marks in this chapter — and what to do instead.
Mistake: Reading the vertical intercept as the constant itself instead of its logarithm.
The plotted quantity is lg y, so the intercept is lg a (or lg k) — a logarithm, not the constant.
Instead: Finish with a = 10^(intercept) (or k = 10^(intercept)), using the given log values.
Mistake: Plotting lg y against lg x for the exponential y = kbˣ.
The exponential linearises with x alone: lg y = (lg b)x + lg k has no lg x term.
Instead: Power laws need logs on both axes; exponentials need a log on the vertical axis only.
Show 8 more mistakesShow fewer mistakes
Mistake: Computing the gradient from two raw data points.
Scattered points each carry error; the line of best fit is the model, not any chord between data points.
Instead: Take two well-separated points on the drawn line itself.
Mistake: Swapping the axes: plotting lg x against lg y.
The gradient would then be 1/n and the "intercept" meaningless for the constants sought.
Instead: The variable on the vertical axis is the one whose expression starts as Y: lg y goes vertical.
Mistake: Forgetting that the gradient of the lg y vs x line is lg b, not b.
b sits in the exponent, so only its logarithm appears as the slope.
Instead: b = 10^(gradient), quoting b to 3 significant figures.
Mistake: Trying to linearise y = x² + 3x with a single log plot.
The two terms share no common factor structure; lg(x² + 3x) does not split.
Instead: Linearisation needs the relation to collect into Y = mX + c form; sums of different powers generally do not.
Mistake: Rounding lg y values to 2 significant figures before plotting.
Premature rounding tilts the drawn line and corrupts the gradient at the third figure.
Instead: Plot to 3 significant figures and read the gradient from well-separated points.
Mistake: Using the model outside the range of the data without comment.
A fitted relation is only trustworthy between the smallest and largest measured values.
Instead: Interpolate freely; label extrapolation as an estimate and say so.
Mistake: Drawing the line of best fit through the origin out of habit.
The intercept carries information (it is lg a or lg k); forcing it to zero destroys that.
Instead: Let the intercept fall where the fit puts it — unless the model genuinely predicts a zero intercept.
Mistake: Leaving the answer as "lg a = 0.30" and stopping.
The question asks for the relationship between x and y in original variables.
Instead: State the final model, e.g. y = 2x^1.5, with constants found.
Check yourself
Quick quiz
15 questions with instant explanations. About 5 minutes.
Question 1 of 3
- Question 1: current question
- Question 2: not answered
- Question 3: not answered
Linearise a power law
For y = axⁿ, plotting lg y against lg x gives a straight line whose gradient is
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Practise
Topical questions
Exam-style practice in three tiers, from recall to data questions.
- 3Foundation
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- 3Challenging
9 questions worth 54 marks, each with a model answer, mark scheme and examiner tips.
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Practice paper
A 30-mark paper set out like a school paper, with a separate mark scheme. Give yourself 40 min.
Question paper
Chapter 3 Practice Paper
30 marks · 40 min · 5 pages
5 multiple-choice questions and structured questions with answer lines. Chapter 3 (G2 linear law: transforming relations to linear form).
Mark scheme
PROChapter 3 Practice Paper — mark scheme
Answers · Marking points · 3 pages
Model answers with a mark for each creditworthy point — for marking at home or in class.
Where this chapter is tested
Each school sets its own scope — check your assessment outline.
Questions
Frequently asked questions
What does Sec 4 A-Math Chapter 3 cover?
It covers the linear-law part of syllabus sub-topic G2 of O-Level 4049 / SEC G3 K341: reducing relationships such as y = axⁿ and y = kbˣ to straight-line form and reading the constants from a line of best fit. The rest of G2 — coordinate geometry and circles — is Sec 3 Chapter 6.
When do I use lg y against lg x, and when lg y against x?
Power laws y = axⁿ straighten with logarithms on both axes (gradient n, intercept lg a); exponentials y = kbˣ straighten with a logarithm on the vertical axis only (gradient lg b, intercept lg k).
How do I recover a constant whose logarithm I read off a graph?
Undo the logarithm with a power of ten: if the intercept is lg a = 0.30, then a = 10^0.30. Exams supply values such as lg 2 = 0.30, lg 3 = 0.48 and lg 5 = 0.70 for this step.
Where does the line of best fit go?
Through the middle of the scattered points, balancing the deviations — not necessarily through any data point, and not forced through the origin. Read the gradient from two far-apart points on the drawn line.
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Chapter 4: Differentiation and Applications