9709 P3 §3.2 — Linearising Data with Logs

💡 Lead In

The Scenario

A creator uploads a video. After 1 hour it has 50 views. After 2 hours: 250 views. After 3 hours: 1250 views. After 4 hours: 6250 views. The creator plots the number of views $V$ against time $t$ — but the graph shoots up so steeply it is almost impossible to read. Then she tries plotting $\ln V$ against $t$ instead. The points fall on a perfect straight line. Why does taking logs turn a wild curve into a tidy line — and how can she read off the growth rate from that line?

The Question

If $V = k \cdot a^t$ describes the views, what equation do you get when you take natural logs of both sides? What does the gradient of the straight line tell you, and what does the intercept tell you?

📖 Key Concepts

Power law:

y = kx^n

Take natural logs of both sides:

\ln y = \ln k + n \ln x

This is a linear equation in the form

Y = c + mX

where

Y = \ln y

X = \ln x

, gradient

= n

, intercept

= \ln k

.
Plot $\ln y$ against $\ln x$ — if the data follows a power law, the points form a straight line. This is called a log-log plot.

k: 2.0

n: 0.5

$y = kx^{n}$ (curved)

$\ln y$ against $\ln x$ (linear)

$\ln y = \ln 2.0 + 0.5 \ln x$

Exponential law:

y = ka^x

Take natural logs of both sides:

\ln y = \ln k + x \ln a

This is linear in

x

: gradient

= \ln a

, intercept

= \ln k

.
Plot $\ln y$ against $x$ — if the data follows an exponential law, the points form a straight line. This is called a semi-log plot.

k: 1.5

a: 2.0

$y = ka^{x}$ (curved)

$\ln y$ against $x$ (linear)

$\ln y = \ln 1.5 + x \ln 2.0$

Summary — reading off $k$ , $n$ , $a$ from a straight-line graph:
Gradient

m

and

y

-intercept

c

of the linearised plot directly give the parameters:
Power law:

n = m

k = e^c

(if using

\ln

) or

k = 10^c

(if using

\log

).
Exponential:

\ln a = m

a = e^m

;

k = e^c

Activity 1 · Group Pattern Detective: The Linearisation Mystery

Detective Briefing

Form groups of 3-4. You are detectives investigating a mystery dataset. The data follows a hidden law. Take logarithms, find the pattern, and crack the case! Work through 3 levels of increasing difficulty.

Detective Score: 40 points

Level 1 The Power Discovery

A mystery dataset has been recovered. Examine the values and identify the hidden law.

x	1	2	3	4	5
y	3	12	27	48	75

Question 1: What type of relationship does this data suggest? ✓ Correct (+5 pts)

Look at how $y$ grows as $x$ increases. When $x$ doubles from 2 to 4, $y$ goes from 12 to 48 (×4). When $x$ triples from 1 to 3, $y$ goes from 3 to 27 (×9). This constant-ratio pattern in the $y$-values points to a power relationship $y = kx^n$.

A linear law:

y = kx + n

A quadratic law:

y = ax^2 + bx + c

An exponential law:

y = ka^x

A power law:

y = kx^n

Question 2: After taking ln of both sides of

y = kx^n

, what do you get? ✓ Correct (+10 pts)

y = \ln k + n \ln x

\ln y = \ln k + n \ln x

\ln y = \ln k \times n \ln x

\ln y = k + n \ln x

Level 1 cracked! The law is

y = 3x^2

$\ln x$	0	0.69	1.10	1.39	1.61
$\ln y$	1.10	2.48	3.30	3.87	4.32

Gradient $= 2 = n$ , intercept $= \ln 3 \approx 1.10 = \ln k$ , so $k = 3$ .

Level 2 The Exponential Enigma

A new dataset has surfaced. This time the pattern is different. Investigate carefully.

t	0	1	2	3	4
P	5	7.4	10.9	16.2	24.0

Question 1: What type of relationship does this data suggest? ✓ Correct (+5 pts)

Compare the ratio of consecutive $P$-values: $\dfrac{7.4}{5} \approx 1.48$, $\dfrac{10.9}{7.4} \approx 1.47$, $\dfrac{16.2}{10.9} \approx 1.49$. Each step multiplies $P$ by almost the same factor — a signature of exponential growth $P = ka^{t}$ (or $P = ke^{ct}$).

A linear law: $P = kt + n$

A logarithmic law: $P = k \ln t$

An exponential law: $P = ka^{t}$ or $P = ke^{ct}$

A power law: $P = kt^{n}$

Question 2: To linearise $P = ka^{t}$, what should you plot? ✓ Correct (+10 pts)

$\ln P$ against $t$ (semi-log plot)

$\ln P$ against $\ln t$ (log-log plot)

$P$ against $\ln t$

$P$ against $t$

Level 2 cracked! The law is

P = 5 \cdot e^{0.4t}

$\ln P = \ln 5 + 0.4t$ . Gradient $= 0.4$ , intercept $= \ln 5 \approx 1.61$ , so $k = 5$ .

Level 3 The Master Challenge

The graph of $\ln y$ against $\ln x$ is a straight line passing through the points $(0,\, 1.6)$ and $(3,\, 3.4)$ .

Question: Find the values of

k

and

n

y = kx^n

. ✓ Correct (+10 pts)

Recall $\ln y = \ln k + n \ln x$. The gradient of the line gives $n$, and the $y$-intercept gives $\ln k$. Gradient $n = \dfrac{3.4 - 1.6}{3 - 0} = 0.6$; the intercept is $1.6$, so $\ln k = 1.6$, meaning $k = e^{1.6} \approx 4.95$.

k = e^{1.6}

n = \dfrac{1.6}{3.4}

k = 1.6

n = 3.4

k = 1.6

n = \dfrac{3.4 - 1.6}{3} = 0.6

k = e^{1.6} \approx 4.95

n = \dfrac{3.4 - 1.6}{3 - 0} = 0.6

Level 3 cracked! Case closed!

k \approx 4.95

n = 0.6

Gradient $= n = \dfrac{3.4 - 1.6}{3 - 0} = 0.6$ . Intercept $= \ln k = 1.6$ , so $k = e^{1.6} \approx 4.95$ .

🎉

Case Solved!

All three levels cracked. You are a Linearisation Detective!

Final Score: 40 points

Summit Reached! 🏽

You conquered the Linearisation Mystery!

✏️ Worked Examples

Worked Example 1 · Group Linearising

A = kb^t

— Semi-Log Plot

Ex 2C Q5 Challenging

The variables $t$ and $A$ satisfy the equation $A = kb^t$ , where $b$ and $k$ are constants.

(i) Show that the graph of $\log A$ against $t$ produces a straight line.

The graph of $\log A$ against $t$ passes through the points $(0, 0.2)$ and $(4, 0.75)$ .

(ii) Find the values of $b$ and $k$ .

Step 1 — Take logs and show linearity (part i)

Starting from $A = kb^t$ , take $\log$ (base 10) of both sides:

$\log A = \log(kb^t) = \log k + \log(b^t) = \log k + t \log b$

This is in the form $Y = c + mt$ where $Y = \log A$ , gradient $= \log b$ , intercept $= \log k$ . So the graph of $\log A$ against $t$ is a straight line. $\square$

Step 2 — Find the gradient and intercept (part ii)

The line passes through $(0, 0.2)$ and $(4, 0.75)$ .

$\text{gradient} = \dfrac{0.75 - 0.2}{4 - 0} = \dfrac{0.55}{4} = 0.1375$

The $y$ -intercept (at $t = 0$ ) is $0.2$ .

Step 3 — Recover

b

and

k

Gradient $= \log b = 0.1375$ , so:

$b = 10^{0.1375} \approx 1.37$

Intercept $= \log k = 0.2$ , so:

$k = 10^{0.2} \approx 1.58$

Final Answer

$b \approx 1.37, \qquad k \approx 1.58$

Worked Example 2 · Individual Linearising

y = Kx^m

— Log-Log Plot

Ex 2D Q13 (9709 P21 Q3 Jun 2011) Challenging

The variables $x$ and $y$ satisfy the equation $y = Kx^m$ , where $K$ and $m$ are constants. The graph of $\ln y$ against $\ln x$ is a straight line passing through the points $(0, 2.0)$ and $(6, 10.2)$ as shown in the diagram. Find the values of $K$ and $m$ , correct to 2 decimal places.

Step 1 — Take ln and identify the linear form

Starting from $y = Kx^m$ , take $\ln$ of both sides:

$\ln y = \ln(Kx^m) = \ln K + \ln(x^m) = \ln K + m \ln x$

This is $Y = c + mX$ where $Y = \ln y$ , $X = \ln x$ , gradient $= m$ , intercept $= \ln K$ .

Step 2 — Find the gradient (gives

m

)

The line passes through $(0, 2.0)$ and $(6, 10.2)$ .

$m = \dfrac{10.2 - 2.0}{6 - 0} = \dfrac{8.2}{6} \approx 1.3667$

$m \approx 1.37 \text{ (2 d.p.)}$

Step 3 — Find the intercept (gives

\ln K

, then

K

)

The $y$ -intercept (at $\ln x = 0$ ) is $2.0$ :

$\ln K = 2.0 \implies K = e^{2.0} \approx 7.389$

$K \approx 7.39 \text{ (2 d.p.)}$

Final Answer

$K \approx 7.39, \qquad m \approx 1.37$

Activity 2 · Individual Sort & Match: Linearisation Pairs

Instructions

Draw a connection between each original equation (left) and its linearised form (right): click a left card, then click its matching right card. Use Reset to clear your connections and try again. When all 4 are connected, they are graded automatically — correct pairs turn green and wrong pairs turn red.

Connected: 0 / 4

Original Equation

y = 4e^{-0.3x}

y = 3x^4

y = 2e^{0.5x}

y = 5x^{-2}

Linearised Form

\ln y = \ln 4 - 0.3x

\ln y = \ln 3 + 4 \ln x

\ln y = \ln 2 + 0.5x

\ln y = \ln 5 - 2 \ln x

⭐

All pairs matched! You are a Linearisation Master!

★★★

Perfect! All 4 pairs matched on the first try.

Concept Quiz — Quick Check

Test your understanding of all four lesson objectives. Each question targets one objective — try each before moving on!

Score: 0 / 4

Q1 (Objective 1) — Power law linearisation

The relationship $y = 3x^2$ is linearised by plotting $\ln y$ against $\ln x$ . What is the gradient of the resulting straight line?

2 \ln 3

3 \ln 2

2

3

$\ln y = \ln(3x^2) = \ln 3 + 2 \ln x$ . The gradient (coefficient of $\ln x$ ) is $2$ .

Q2 (Objective 2) — Exponential law linearisation

For the model $y = 5 \cdot 2^x$ , what is the $y$ -intercept of the graph of $\ln y$ against $x$ ?

5

2

\ln 2

\ln 5

$\ln y = \ln 5 + x \ln 2$ . At $x = 0$ , $\ln y = \ln 5$ . So the intercept is $\ln 5 \approx 1.61$ .

Q3 (Objective 3) — Recovering

k

and

n

A log-log plot of $\ln y$ against $\ln x$ has gradient $0.5$ and intercept $1$ . If $y = kx^n$ , what are $k$ and $n$ ?

k = e^{0.5}, \; n = 1

k = e, \; n = 0.5

k = 0.5, \; n = 1

k = 1, \; n = 0.5

Gradient $= n = 0.5$ . Intercept $= \ln k = 1$ , so $k = e^1 = e \approx 2.72$ .

Q4 (Objective 4) — Choosing the right plot

Data is believed to follow $y = ka^x$ . Which plot produces a straight line?

\ln y

against

\ln x

(log-log)

\ln y

against

x

(semi-log)

y

against

x

y

against

\ln x

$\ln y = \ln k + x \ln a$ is linear in $x$ (not $\ln x$ ). So plotting $\ln y$ against $x$ (a semi-log plot) gives a straight line. A log-log plot ( $\ln y$ vs $\ln x$ ) is used for power laws $y = kx^n$ .

🎉

You scored 0 / 4

More Practice

Exercise 2C Q1 Standard

The planet Saturn has many moons. The table below gives the mean radius of orbit and the time taken to complete one orbit for five of the best-known of them.

$\begin{array}{|l|c|c|c|c|c|} \hline \text{Moon} & \text{Tethys} & \text{Dione} & \text{Rhea} & \text{Titan} & \text{Iapetus} \\ \hline \text{Radius } R \, (\times 10^5 \text{ km}) & 2.9 & 3.8 & 5.3 & 12.2 & 35.6 \\ \hline \text{Period } T \, (\text{days}) & 1.9 & 2.7 & 4.5 & 15.9 & 79.3 \\ \hline \end{array}$

It is believed that the relationship between $R$ and $T$ is of the form $R = kT^n$ .

(i) How can this be tested by plotting $\log R$ against $\log T$ ?

(ii) Make out a table of values of $\log R$ and $\log T$ and draw the graph.

(iii) Use your graph to estimate the values of $k$ and $n$ .

Part (i) — Testing the model

Take logs of $R = kT^n$ :

$\log R = \log k + n \log T$

This is linear in $\log T$ , so plotting $\log R$ against $\log T$ should give a straight line if the model holds. The gradient gives $n$ and the intercept gives $\log k$ .

Part (ii) — Table of log values

$\begin{array}{|c|c|c|c|c|c|} \hline \log T & 0.279 & 0.431 & 0.653 & 1.201 & 1.899 \\ \hline \log R & 0.462 & 0.580 & 0.724 & 1.086 & 1.551 \\ \hline \end{array}$

Plot these points on graph paper; they should lie approximately on a straight line, as shown below:

Scatter graph of log R against log T with the line of best fit

Part (iii) — Estimate

k

and

n

Using two points on the line of best fit, e.g. $(\log T, \log R) = (0.279, 0.462)$ and $(1.899, 1.551)$ :

$n = \dfrac{1.551 - 0.462}{1.899 - 0.279} = \dfrac{1.089}{1.620} \approx 0.67$

Intercept: $\log k \approx 0.462 - 0.67 \times 0.279 \approx 0.275$ , so $k \approx 10^{0.275} \approx 1.88$ (in units of $10^5$ km when $T$ is in days).

$n \approx 0.67 \;(=\tfrac{2}{3}\text{, Kepler's law}), \qquad k \approx 1.9$

Exercise 2C Q2 Standard

The table below shows the area, $A \, \mathrm{cm^2}$ , occupied by a patch of mould at time $t$ days since measurements were started.

$\begin{array}{|l|c|c|c|c|c|c|} \hline t & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline A & 0.9 & 1.3 & 1.8 & 2.5 & 3.5 & 5.2 \\ \hline \end{array}$

It is believed that $A$ may be modelled by a relationship of the form $A = kb^t$ .

(i) Show that the model may be written as $\log A = t\log b + \log k$ .

(ii) What graph must be plotted to test this model?

(iii) Plot the graph and use it to estimate the values of $b$ and $k$ .

(iv) (a) Estimate the time when the area of mould was $2 \, \mathrm{cm^2}$ . (b) Estimate the area of the mould after 3.5 days.

(v) How is this sort of growth pattern described?

Hint

For part (i), take $\log$ of both sides of $A = kb^t$ and use the laws of logs. For part (iii), plot $\log A$ against $t$ — the gradient is $\log b$ and the intercept is $\log k$ .

Part (i) — Show linear form

$\log A = \log(kb^t) = \log k + t \log b$

This is $\log A = (\log b) \cdot t + \log k$ , a linear equation in $t$ . $\square$

Part (ii) — Which graph to plot

Plot $\log A$ against $t$ (a semi-log plot). If the points lie on a straight line, the model $A = kb^t$ is valid.

Part (iii) — Estimate

b

and

k

Log values: $\log A \approx -0.046, 0.114, 0.255, 0.398, 0.544, 0.716$ for $t = 0, 1, 2, 3, 4, 5$ .

Scatter graph of log A against t with the line of best fit

Gradient $\approx \dfrac{0.716 - (-0.046)}{5 - 0} = \dfrac{0.762}{5} \approx 0.152$ , so $\log b \approx 0.152$ and $b \approx 10^{0.152} \approx 1.42$ .

Intercept $\approx -0.046$ , so $\log k \approx -0.046$ and $k \approx 10^{-0.046} \approx 0.90$ .

$b \approx 1.4, \qquad k \approx 0.9$

Part (iv) — Estimates

(a) Set $A = 2$ : $\log 2 = t \log b + \log k \Rightarrow 0.301 = 0.152t - 0.046 \Rightarrow t \approx \dfrac{0.347}{0.152} \approx 2.3$ days.

(b) At $t = 3.5$ : $A = 0.9 \times 1.42^{3.5} \approx 0.9 \times 3.43 \approx 3.1 \, \mathrm{cm^2}$ .

Part (v) — Growth pattern

This is exponential growth — the area increases by a constant ratio ( $\times b$ ) each day.

Exercise 2D Q11 Challenging

The variables $x$ and $y$ satisfy the equation $y = A(b^x)$ , where $A$ and $b$ are constants. The graph of $\ln y$ against $x$ is a straight line passing through the points $(0, 1.3)$ and $(1.6, 0.9)$ as shown in the diagram. Find the values of $A$ and $b$ , correct to 2 decimal places.

Hint

Take $\ln$ of $y = A b^x$ : $\ln y = \ln A + x \ln b$ . The gradient of the $\ln y$ vs $x$ line is $\ln b$ , and the intercept is $\ln A$ .

Step 1 — Linearise

$\ln y = \ln(A b^x) = \ln A + x \ln b$

Gradient $= \ln b$ , intercept $= \ln A$ .

Step 2 — Find the gradient (gives

b

)

$\text{gradient} = \dfrac{0.9 - 1.3}{1.6 - 0} = \dfrac{-0.4}{1.6} = -0.25$

$\ln b = -0.25 \implies b = e^{-0.25} \approx 0.7788$

$b \approx 0.78 \text{ (2 d.p.)}$

Step 3 — Find the intercept (gives

A

)

At $x = 0$ , $\ln y = 1.3$ :

$\ln A = 1.3 \implies A = e^{1.3} \approx 3.6693$

$A \approx 3.67 \text{ (2 d.p.)}$

Final Answer

$A \approx 3.67, \qquad b \approx 0.78$

Exercise 2C Q9 Advanced

The variables $x$ and $y$ satisfy the relation $3^y = 4^{x+2}$ .

(i) By taking logarithms, show that the graph of $y$ against $x$ is a straight line. Find the exact value of the gradient of this line.

(ii) Calculate the $x$ coordinate of the point of intersection of this line with the line $y = 2x$ , giving your answer correct to 2 decimal places.

Hint 1

Take $\ln$ (or $\log$ ) of both sides. Use $\ln(a^b) = b \ln a$ to expand. The result will be $y = (\text{constant}) \cdot x + (\text{constant})$ — a linear equation in $x$ , so the graph of $y$ against $x$ is already a straight line (no need to take logs of $y$ again).

Part (i) — Show linearity and find gradient

Take $\ln$ of both sides of $3^y = 4^{x+2}$ :

$y \ln 3 = (x + 2) \ln 4$

$y = \dfrac{\ln 4}{\ln 3} x + \dfrac{2 \ln 4}{\ln 3}$

This is in the form $y = mx + c$ , so the graph of $y$ against $x$ is a straight line. $\square$

$\text{gradient} = \dfrac{\ln 4}{\ln 3} \text{ (exact)}$

Part (ii) — Intersection with

y = 2x

Set the two expressions for $y$ equal:

$2x = \dfrac{\ln 4}{\ln 3} x + \dfrac{2 \ln 4}{\ln 3}$

$x \left(2 - \dfrac{\ln 4}{\ln 3}\right) = \dfrac{2 \ln 4}{\ln 3}$

$x = \dfrac{2 \ln 4}{2 \ln 3 - \ln 4}$

Using $\ln 4 = 2 \ln 2$ :

$x = \dfrac{4 \ln 2}{2 \ln 3 - 2 \ln 2} = \dfrac{2 \ln 2}{\ln 3 - \ln 2}$

$x \approx \dfrac{2 \times 0.6931}{1.0986 - 0.6931} = \dfrac{1.3863}{0.4055} \approx 3.4190$

$x \approx 3.42 \text{ (2 d.p.)}$

Check

At $x = 3.42$ : $y = 2(3.42) = 6.84$ . Check $3^{6.84} \approx 1830.5$ and $4^{3.42+2} = 4^{5.42} \approx 1830.5$ . $\checkmark$

🎫

Exit Ticket

Check the learning objectives you feel confident about:

I can linearise

y = kx^n

by taking logs to get

\ln y = \ln k + n \ln x

and plot

\ln y

against

\ln x

. I can linearise

y = ka^x

by taking logs to get

\ln y = \ln k + x \ln a

and plot

\ln y

against

x

. I can recover

k

n

, and

a

from the gradient and intercept of a linearised log-graph. I can decide whether to use a log-log plot or a semi-log plot depending on the model.