Logarithmic Graphs

This lesson covers how logarithmic graphs are used to determine unknown constants in relationships of the form y = axⁿ and y = abˣ. This technique of "reducing to linear form" is a key part of the Edexcel 9MA0 specification and appears regularly in exam papers.

Why Use Logarithmic Graphs?

When data follows a power law (y = axⁿ) or an exponential law (y = abˣ), plotting the raw data gives a curve. It is difficult to read off the constants a and n (or a and b) from a curve.

By taking logarithms, we can transform the relationship into a straight line of the form Y = mX + c, where the gradient and intercept give us the unknown constants.

Case 1: y = axⁿ (Power Model)

Take log₁₀ of both sides:

log(y) = log(a × xⁿ) = log(a) + log(xⁿ) = log(a) + n × log(x)

This has the form:

log(y) = n × log(x) + log(a)

Comparing with Y = mX + c:

Plot log(y) on the vertical axis and log(x) on the horizontal axis
Gradient = n (the power)
y-intercept = log(a), so a = 10^(intercept)

Example: Data suggests y = axⁿ. Given log(x) and log(y) values, the line of best fit has gradient 2.5 and y-intercept 0.7.

n = 2.5
log(a) = 0.7, so a = 10^0.7 ≈ 5.01
The relationship is y = 5.01x^2.5

Case 2: y = abˣ (Exponential Model)

Take log₁₀ of both sides:

log(y) = log(a × bˣ) = log(a) + log(bˣ) = log(a) + x × log(b)

This has the form:

log(y) = log(b) × x + log(a)

Comparing with Y = mX + c:

Plot log(y) on the vertical axis and x on the horizontal axis (note: x itself, not log(x))
Gradient = log(b), so b = 10^(gradient)
y-intercept = log(a), so a = 10^(intercept)

Example: Data suggests y = abˣ. The graph of log(y) against x gives a straight line with gradient 0.3 and y-intercept 1.2.

log(b) = 0.3, so b = 10^0.3 ≈ 1.995 ≈ 2
log(a) = 1.2, so a = 10^1.2 ≈ 15.85
The relationship is y ≈ 15.85 × 2ˣ

Using ln Instead of log₁₀

You can use natural logarithms instead. The method is identical:

For y = axⁿ: ln(y) = n × ln(x) + ln(a) For y = abˣ: ln(y) = ln(b) × x + ln(a)

For the exponential model y = Aeᵏˣ (base e):

ln(y) = kx + ln(A)
Plot ln(y) against x
Gradient = k
Intercept = ln(A), so A = e^(intercept)

The choice of log₁₀ or ln makes no difference to the gradient and intercept values for the power model. For the exponential model, using ln is simpler when the base is e.

How to Read the Graph in an Exam

You will often be given a graph with:

Two labelled points on the line, e.g. (2, 3.4) and (5, 5.5)
Or the equation of the line of best fit

Step 1: Calculate the gradient: m = (y₂ - y₁)/(x₂ - x₁)

Step 2: Find the intercept using y = mx + c and one of the points.

Step 3: Interpret m and c according to the model (as above).

Example: The line passes through (1, 2.5) and (4, 4.0) on a graph of log(y) vs log(x).

Gradient = (4.0 - 2.5)/(4 - 1) = 1.5/3 = 0.5
Using (1, 2.5): 2.5 = 0.5(1) + c, so c = 2
The model is y = axⁿ where n = 0.5 and log(a) = 2, so a = 100
y = 100x^0.5 = 100√x

Distinguishing Between the Two Models

Model	Log graph axes	Evidence of correct model
y = axⁿ	log(y) vs log(x)	Straight line
y = abˣ	log(y) vs x	Straight line

If log(y) vs log(x) gives a straight line, the data follows a power model. If log(y) vs x gives a straight line, the data follows an exponential model.

Exam Tip: Pay careful attention to the axis labels. The axes tell you which model is being used. If the horizontal axis is log(x), it is a power model. If it is just x, it is an exponential model.

Worked Exam-Style Problem

The table shows experimental data:

x	2	4	6	8
y	7.4	54.6	403	2981

Test whether y = abˣ by plotting log₁₀(y) against x.

log₁₀(y) values: 0.869, 1.737, 2.605, 3.474

Plotting these against x gives points that lie on a straight line.

Gradient = (3.474 - 0.869)/(8 - 2) = 2.605/6 ≈ 0.434

Using (2, 0.869): 0.869 = 0.434(2) + c, so c = 0.001 ≈ 0

Therefore: log(b) = 0.434, so b ≈ 2.72 ≈ e, and log(a) ≈ 0, so a ≈ 1.

The model is y ≈ eˣ (approximately).

Summary

For y = axⁿ: plot log(y) against log(x); gradient = n, intercept = log(a).
For y = abˣ: plot log(y) against x; gradient = log(b), intercept = log(a).
Read the gradient and intercept from the graph, then convert back to find a, n, or b.
Always check which axes are used to determine the correct model.

A-Level Deep Dive: Logarithmic Graphs

Spec mapping

Edexcel 9MA0-01 specification section 6 — Exponentials and logarithms, sub-strand on linearising data using logarithms covers logarithmic graphs to estimate parameters in relationships of the form $y = ax^n$ and $y = kb^x$ , given data for $x$ and $y$ (refer to the official specification document for exact wording). This sits at the practical, modelling end of section 6 and is the part most likely to appear in Paper 1 modelling questions as a 6-to-10-mark structured item. The technique is examined synoptically against statistics (line of best fit and linear regression in 9MA0-03 section 3), against modelling assumptions and refinement language (cross-paper), against experimental physics analysis (real-world contexts) and against coordinate geometry (gradient and intercept of $y = mx + c$ ). The Edexcel formula booklet does not list the linearisation derivations — they must be reproduced from first principles using the laws $\log(ab) = \log a + \log b$ and $\log(a^n) = n \log a$ .

Worked example with full mark scheme

Question (8 marks):

A scientist measures a quantity $y$ at six values of $x$ and obtains the following data:

$x$	2	4	6	8	10	12
$y$	5.66	16.0	29.4	45.3	63.2	83.1

The scientist believes $y$ is related to $x$ by either $y = A x^n$ or $y = A b^x$ for constants $A$ , $n$ or $b$ .

(a) By taking logarithms of both sides of each candidate model, state the equation of the straight line that should result for each, identifying the gradient and intercept in terms of the unknown constants. (2)

(b) Using the data, decide which model is appropriate. Justify your decision. (2)

Solution with mark scheme:

(a) Step 1 — linearise the power model.

Taking $\log_{10}$ of both sides of $y = A x^n$ :

$\log y = \log A + n \log x$

This is linear in $\log x$ with gradient $n$ and intercept $\log A$ , plotted as $\log y$ against $\log x$ (a log-log plot).

M1 — correct application of $\log(AB) = \log A + \log B$ and $\log(x^n) = n \log x$ .

Step 2 — linearise the exponential model.

Taking $\log_{10}$ of both sides of $y = A b^x$ :

$\log y = \log A + x \log b$

This is linear in $x$ (not $\log x$ ) with gradient $\log b$ and intercept $\log A$ , plotted as $\log y$ against $x$ (a semi-log plot).

A1 — both linearisations stated correctly with gradient and intercept identified.

(b) Step 1 — compute log values for both candidate plots.

$x$	$\log x$	$\log y$
2	0.301	0.753
4	0.602	1.204
6	0.778	1.468
8	0.903	1.656
10	1.000	1.801
12	1.079	1.920

M1 — at least four $\log x$ and $\log y$ values computed correctly.

Step 2 — compare linearity.

For the log-log plot ( $\log y$ vs $\log x$ ): the ratios $\Delta(\log y) / \Delta(\log x)$ between successive points are 1.50, 1.50, 1.50, 1.49, 1.51 — essentially constant. This is a straight line.

For the semi-log plot ( $\log y$ vs $x$ ): the gradient $\Delta(\log y) / \Delta x$ over equal $x$ -intervals of 2 gives 0.226, 0.132, 0.094, 0.073, 0.060 — clearly not constant. So the semi-log plot is curved.

The data fits the power model $y = A x^n$ .

A1 — correct conclusion with quantitative justification (constant gradient on log-log, non-constant on semi-log).

Using the endpoints $(\log x, \log y) = (0.301, 0.753)$ and $(1.079, 1.920)$ :

$n = \dfrac{1.920 - 0.753}{1.079 - 0.301} = \dfrac{1.167}{0.778} = 1.50$

M1 — correct gradient method on the log-log line.

A1 — $n = 1.5$ (2 s.f.).

Step 2 — find the intercept $\log A$ .

Using $(\log x, \log y) = (1.000, 1.801)$ :

$\log A = \log y - n \log x = 1.801 - 1.5 \times 1.000 = 0.301$

So $A = 10^{0.301} = 2.00$ (2 s.f.).

M1 — correct rearrangement and back-substitution to recover $A$ from $\log A$ .

A1 — $A = 2.0$ (2 s.f.). The model is $y = 2.0 x^{1.5}$ .

Total: 8 marks (M4 A4, split as shown).

Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): A biologist models the mass $m$ (in grams) of a culture as a function of time $t$ (in hours). She suspects $m = A b^t$ for constants $A > 0$ and $b > 0$ , and produces a plot of $\log_{10} m$ against $t$ . The line of best fit has gradient $0.18$ and $\log_{10} m$ -intercept $1.30$ .

(a) State, with reasoning, why a plot of $\log_{10} m$ against $t$ (rather than against $\log_{10} t$ ) is appropriate for this model. (2)

(b) Find the values of $A$ and $b$ , each to 2 significant figures. (3)

Mark scheme decomposition by AO:

(a)

M1 (AO2.1) — taking logs of $m = A b^t$ to obtain $\log m = \log A + t \log b$ , identifying that this is linear in $t$ (not in $\log t$ ).
A1 (AO2.4) — explicit reasoning that the exponential structure $b^t$ produces a term $t \log b$ that is linear in $t$ directly, hence semi-log axes are appropriate.

(b)

M1 (AO1.1b) — $\log b = 0.18$ , so $b = 10^{0.18}$ .
A1 (AO1.1b) — $b = 1.51$ to 2 s.f., correctly rounded.
A1 (AO1.1b) — $\log A = 1.30$ , so $A = 10^{1.30} = 20$ (2 s.f.).

(c)

B1 (AO3.4) — substituting $t = 10$ into the recovered model $m = 20 \times 1.51^{10}\approx 20 \times 60 = 1200$ g (or equivalently $\log m = 1.30 + 0.18 \times 10 = 3.10$ , so $m = 10^{3.10} \approx 1260$ g; either route earns the mark).

Total: 6 marks split AO1 = 3, AO2 = 2, AO3 = 1. The AO2 weight is unusual for a section-6 procedural question, but typical for a modelling-flavoured item — Edexcel rewards explicit identification of why the chosen axes work.

Synoptic links

Connects to:

9MA0-03 Statistics, section 3 (Statistical regression): the line of best fit on a log-log or semi-log scatter is a linear regression problem. Edexcel's regression line $y = a + bx$ with sum-of-squares formulae for $a$ and $b$ applies directly to $(\log x, \log y)$ or $(x, \log y)$ pairs. The product-moment correlation coefficient $r$ on the transformed data measures how well the model linearises the relationship — a value of $r > 0.99$ on log-log axes is strong evidence for a power-law relationship.
Modelling and modelling refinement (cross-paper): Edexcel's "modelling cycle" (formulate → solve → interpret → refine) sits behind every linearisation question. The decision to plot log-log versus semi-log is itself a modelling choice, and the residual pattern after fitting tells you whether to refine the model further (e.g. to $y = A x^n + c$ ).
Experimental physics — real-world data analysis: Kepler's third law ( $T^2 \propto r^3$ , equivalently $T = A r^{3/2}$ ), Stefan-Boltzmann's law ( $P = \sigma T^4$ ), Hooke's-law deviations (power-law stress-strain) and radioactive decay ( $N = N_0 e^{-\lambda t}$ ) are all detected and parameterised by exactly the techniques in this lesson. A-Level Physics uses this every term.
Coordinate geometry (sections 1 and 3): the gradient-intercept form $y = mx + c$ is the engine here — recognising that $\log y = (n) \log x + (\log A)$ is structurally identical to $y = mx + c$ is the cognitive step. Reading $m$ off a graph is GCSE; reading $\log A$ off a log-log plot and exponentiating to recover $A$ is A-Level.
Logarithm laws (earlier in section 6): the entire technique rests on $\log(ab) = \log a + \log b$ and $\log(a^n) = n \log a$ . Without absolute fluency on these, the linearisation cannot even be stated, let alone applied.

Mark-scheme literacy

Logarithmic-graph questions on 9MA0 split AO marks across three pillars, with a noticeably higher AO3 share than most section-6 procedural items:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	40–55%	Taking logs of both sides correctly, computing log values, reading gradient and intercept off a line, exponentiating $\log A$ to recover $A$
AO2 (reasoning / interpretation)	25–35%	Stating why log-log linearises a power model (versus semi-log for exponential), justifying the choice of axes from data, interpreting gradient as $n$ or $\log b$
AO3 (problem-solving / modelling)	15–25%	Choosing between competing models from data, making predictions, refining the model after residual inspection

Examiner-rewarded phrasing: "taking logs of both sides gives $\log y = n \log x + \log A$ "; "the gradient gives $\log b$ , so $b = 10^{\text{gradient}}$ "; "the $\log y$ -intercept gives $\log A$ , so $A = 10^{\text{intercept}}$ "; "since the data is linear on log-log axes, the model is $y = A x^n$ ". Phrases that lose marks: "the gradient gives $b$ " (it gives $\log b$ , not $b$ ); "the intercept gives $A$ " (it gives $\log A$ ); writing $A = \text{intercept}$ without the exponentiation step.

A specific Edexcel pattern: when the question says "use a logarithmic plot to estimate $n$ ", the M-marks reward the linearisation step explicitly written down, not the final value. Writing the value $n = 1.5$ with no derivation (even if correct) typically earns one A1 and zero M-marks — the method has not been demonstrated.

Grade-band model answers

3-mark question

Question: State, with brief justification, which transformation linearises the model $y = A x^n$ for unknown constants $A > 0$ and $n$ , and identify the gradient and intercept of the resulting straight line.

Grade C response (~210 words):

To linearise $y = A x^n$ I take logs of both sides:

$\log y = \log A + n \log x$ .

This is in the form $Y = mX + c$ where $Y = \log y$ , $X = \log x$ , $m = n$ and $c = \log A$ .

So plotting $\log y$ on the vertical axis against $\log x$ on the horizontal axis gives a straight line. The gradient is $n$ and the intercept is $\log A$ .

This is called a log-log plot because both axes are logarithmic.

Logarithmic Graphs

Logarithmic Graphs

Why Use Logarithmic Graphs?

Case 1: y = axⁿ (Power Model)

Case 2: y = abˣ (Exponential Model)

Using ln Instead of log₁₀

How to Read the Graph in an Exam

Distinguishing Between the Two Models

Worked Exam-Style Problem

Summary

A-Level Deep Dive: Logarithmic Graphs

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

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