Learn9709Paper 2 · Pure Mathematics 2Logarithmic and exponential functionsExponential models and linear forms

Logarithmic and exponential functions · Subtopic 5 of 5

Exponential Models and Linear Forms

Using logarithms in the real world: modelling growth and decay with $Ae^{kt}$ , and turning curves like $y=kx^n$ and $y=ka^x$ into straight lines so their constants can be read from a graph.

Cambridge 9709 Paper 2Cambridge 9709 Paper 3Edexcel YMA01 WMA13Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1≈ 45 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Use models of the form $y=Ae^{kt}$ for growth ( $k>0$ ) and decay ( $k<0$ ).
Find constants in a model from given data, and find when a quantity reaches a value.
Take logs to turn $y=kx^n$ into $\ln y=\ln k+n\ln x$ and $y=ka^x$ into $\ln y=\ln k+x\ln a$ .
Read $k$ , $n$ or $a$ from the gradient and intercept of a straight-line graph.

Pause and think

Take logs of both sides of $y=kx^n$ . What do you notice about the result if you call $Y=\ln y$ and $X=\ln x$ ?

Section 2 of 14

Core idea

Start hereStraightening a curve

A biologist measures bacteria every hour and plots them: a curve shooting upward. Is it really exponential? It's hard to tell by eye. Take logs of the counts and replot: if the points line up, it's exponential, and the line's gradient and intercept give the growth constants.

The big idea

$y=kb^x$ becomes $\ln y=x\ln b+\ln k$ (plot $\ln y$ against $x$ ). $y=ax^n$ becomes $\ln y=n\ln x+\ln a$ (plot $\ln y$ against $\ln x$ ). Read constants from the gradient and intercept. Growth/decay models: $P=P_0e^{kt}$ .

1. Exponential models

Worked exampleBacteria doubling every 3 hours

A culture of $500$ doubles every $3$ h. Model $P=500e^{kt}$ . Find $k$ , $P$ after 10 h, and when $P=20\,000$ .

Goal: $k$ , a value, a time.
Strategy: Use the doubling fact to find $k$ .

$1000=500e^{3k}\Rightarrow e^{3k}=2\Rightarrow k=\tfrac{\ln2}3=0.231$
Doubling.
$P(10)=500e^{2.31}\approx5040$
Substitute.
$e^{0.231t}=40\Rightarrow t=\frac{\ln40}{0.231}=16.0\text{ h}$
Solve.

Answer

k=0.231

;

5040

;

16.0

✓ Check

10

h is

3\frac13

doublings:

500\times2^{3.33}=5040

✓.

2. Linearising

Two transformations

$y=kb^x\Rightarrow\ln y=\ln k+x\ln b$
Logs; product and power laws.
$\underbrace{\ln y}_{Y}=\underbrace{\ln b}_{m}\,x+\underbrace{\ln k}_{c}$
Straight line in $(x,\ln y)$ .
$y=ax^n\Rightarrow\ln y=n\ln x+\ln a$
Straight line in $(\ln x,\ln y)$ .

The same data before and after taking logs

The curve becomes a straight line when you plot

\ln y

. Its gradient is

\ln1.3\approx0.26

and intercept

\ln3\approx1.10

Worked exampleFrom a log–log line to $y=ax^n$

$\ln y$ against $\ln x$ is a line through $(1,2.5)$ and $(3,6.5)$ . Find $a$ and $n$ .

Goal: Constants.
Strategy: Gradient = $n$ ; intercept = $\ln a$ .

$n=\frac{6.5-2.5}{3-1}=2$
Gradient.
$2.5=2(1)+\ln a\Rightarrow\ln a=0.5$
Substitute a point.
$a=e^{0.5}=1.65$
Un-log.

Answer

y=1.65x^2

✓ Check

x=e

\ln y=0.5+2=2.5

✓ (point

(1,2.5)

Which graph makes a straight line?

Drawing diagram…

A straight line confirms the model; its gradient and intercept give the constants.

Pause and think

Which graph is linear for $y=3\times5^x$ ?

Summary

$P_0e^{kt}$ models.
Exponential: log–linear; power: log–log.

Linearise: Transform so data lie on a straight line.

HistoryMalthus and population. In 1798 Thomas Malthus argued that populations grow exponentially while food supplies grow only steadily, so famine was inevitable. Exponential models still underpin predictions about populations, epidemics and money, though real growth always slows eventually.

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