Mathematics 9709 · Pure Mathematics 3

Straightening curves: turn y = kxⁿ or y = Ae^(bx) into Y = mX + c

This is the most common P3 type: a 4- or 5-mark question in nearly every series, usually dressed up as a real model (bacteria, decay, a data table).
A curved law like $y=kx^n$ or $y=Ae^{bx}$ won't plot as a straight line. Take $\ln$ and it goes straight.
That works because $\ln$ turns multiplying and powers into adding and multiplying, which is the shape of $Y=mX+c$ .

Take $\ln$ of both sides, then line it up with $Y=mX+c$ . $X$ and $Y$ hold only the original variables. $m$ (gradient) and $c$ (intercept) hold only the unknown constants. (textbook §2.8, WE 2.15)

First, see the model: growth vs decay

Drag the rate $b$ below and watch the two things that never move. They are exactly what the linearising method reads off.

rate k

y = 2 e^0.50x·growth·doubles every 1.39 in x

Every curve goes through $(0,A)$ because $e^{0}=1$ . So the intercept is the constant $A$ (after you take $\ln$ ). The rate $b$ only sets how fast it moves: $b>0$ grows, $b<0$ decays, and $b$ is the gradient. In short: gradient is the rate, intercept holds the starting constant.

The two model shapes and their recipes

Model	Take ln →	Y vs X	Read off
$y=k\,x^n$	$\ln y=n\ln x+\ln k$	$\ln y$ vs $\ln x$	gradient $=n$ , intercept $=\ln k$
$y=A\,e^{bx}$	$\ln y=b\,x+\ln A$	$\ln y$ vs $x$	gradient $=b$ , intercept $=\ln A$

To get the constant back, undo the $\ln$ . From a $\ln$ -graph, $k=e^{\text{intercept}}$ (and $A=e^{\text{intercept}}$ ). Stopping at $\ln k$ is only half an answer.

One quick way to tell which shape you're in

Look at what sits on the bottom axis. $\ln x$ on the bottom → you are straightening a power law $y=kx^n$ (the gradient is the power $n$ ). Plain $x$ on the bottom → you are straightening an exponential $y=Ae^{bx}$ (the gradient is the rate $b$ ). The up axis is $\ln y$ in both cases.

The exponential model, step by step

Worked example

$P=ae^{kt}$ : $\ln P$ against $t$ has gradient $\tfrac{1}{20}$ , cutting the axis at $(0,3)$

$P$ is a bacteria count at time $t$ hours. Find $k$ , and $a$ to 2 s.f.

Take $\ln$ : $\ln P=\ln a+kt$ , i.e. $\ln P=k\,t+\ln a$ .
Match to $Y=mX+c$ with $Y=\ln P$ , $X=t$ : gradient $=k$ , intercept $=\ln a$ .
So $k=\dfrac{1}{20}$ straight off the gradient.
Intercept gives $\ln a=3$ , so recover with $a=e^3\approx20$ .

Answer

k=\tfrac{1}{20},\quad a=e^3\approx20

(9709/33 Oct/Nov 2024 Q3(a))

Your turn— tap to reveal the worked answer (9709/33 Oct/Nov 2024 Q3(b))

With $P=e^3\,e^{t/20}$ , find how long it takes $P$ to double. Doubling means $e^{t/20}=2$ (the $e^3$ cancels). Take $\ln$ : $\dfrac{t}{20}=\ln 2$ .

Answer

t=20\ln 2\approx14\ \text{hours}

When you're given two points

Often you just get two points on the $\ln y$ -against- $x$ line.
The gradient between the two points is the constant. Then put one point back in to get the intercept.

Worked example

$ky=e^{cx}$ : line $\ln y$ vs $x$ through $(2.80,\,0.372)$ and $(5.10,\,2.21)$

Find $k$ and $c$ to 2 s.f.

Rearrange first: $ky=e^{cx}$ gives $\ln k+\ln y=cx$ , so $\ln y=cx-\ln k$ — gradient $c$ , intercept $-\ln k$ .
Gradient from the points: $c=\dfrac{2.21-0.372}{5.10-2.80}=0.80$ .
Sub a point for the intercept: $0.372=0.80(2.80)-\ln k\Rightarrow\ln k=1.87$ .
Recover: $k=e^{1.87}\approx6.5$ .

Answer

c=0.80,\quad k=6.5

(9709/33 May/Jun 2024 Q2)

Worked example

$y^2=Ae^{kx}$ : line $\ln y$ vs $x$ through $(1.5,\,1.2)$ and $(5.24,\,2.7)$

The hidden-power version. Find $A$ and $k$ to 2 d.p., and watch the $y^2$ .

Take $\ln$ : $\ln(y^2)=\ln A+kx$ , and $\ln(y^2)=2\ln y$ , so $2\ln y=kx+\ln A$ .
Divide by 2 to match the plotted axis $\ln y$ : $\ln y=\tfrac{k}{2}\,x+\tfrac12\ln A$ .
Gradient of the plotted line: $\dfrac{2.7-1.2}{5.24-1.5}=0.40$ , and that equals $\tfrac{k}{2}$ , so $k=0.80$ .
Intercept (sub a point): $1.2=0.40(1.5)+\tfrac12\ln A\Rightarrow\tfrac12\ln A=0.60$ , so $\ln A=1.20$ and $A=e^{1.2}\approx3.32$ .

Answer

A\approx3.32,\quad k=0.80

(9709/32 May/Jun 2020 Q2)

Read the axes — they tell you which constant is which

The biggest trap isn't the algebra. It is mis-reading what is plotted against what, so read the axis labels to see which constant is the gradient and which is the intercept.

Axes given	Means the model is	Watch for
$\ln y$ vs $x$	exponential $y=Ae^{bx}$	gradient is the rate
$\ln y$ vs $\ln x$	power $y=kx^n$	gradient is the power
$y$ vs $\ln x$	roles flipped	don't assume the standard shape

Common mistake

Three classic lost marks: (1) mixing up which constant is gradient and which is intercept; (2) forgetting to do

e^{\text{intercept}}

and leaving the answer as

\ln k

; (3) reading

y

\ln x

as if it were

\ln y

\ln x

, which swaps the roles completely.

Harder versions: when BOTH sides need a log first

Some questions give $a^y=b^x$ . Take $\ln$ of both sides: $y\ln a=x\ln b$ . You only get $\ln y$ -against- $x$ after one more rearrange, so don't just assume the standard shape. (9709/32 Oct/Nov 2024 Q6)

And the power-law version with two unknowns: $x^n y^2=C$ . Take $\ln$ : $n\ln x+2\ln y=\ln C$ , so $\ln y=-\tfrac{n}{2}\ln x+\tfrac12\ln C$ . A factor of $2$ is added, just like it was with $y^2$ above. (9709/32 Feb/Mar 2022 Q3)

Previouse^x and ln x: inverse twins, graphs, and the e/ln cancel NextCosecant, secant and cotangent

Straightening curves: turn y = kxⁿ or y = Ae^(bx) into Y = mX + c

This is the most common P3 type: a 4- or 5-mark question in nearly every series, usually dressed up as a real model (bacteria, decay, a data table).
A curved law like $y=kx^n$ or $y=Ae^{bx}$ won't plot as a straight line. Take $\ln$ and it goes straight.
That works because $\ln$ turns multiplying and powers into adding and multiplying, which is the shape of $Y=mX+c$ .

First, see the model: growth vs decay

Drag the rate $b$ below and watch the two things that never move. They are exactly what the linearising method reads off.

rate k

y = 2 e^0.50x·growth·doubles every 1.39 in x

The two model shapes and their recipes

Model	Take ln →	Y vs X	Read off
$y=k\,x^n$	$\ln y=n\ln x+\ln k$	$\ln y$ vs $\ln x$	gradient $=n$ , intercept $=\ln k$
$y=A\,e^{bx}$	$\ln y=b\,x+\ln A$	$\ln y$ vs $x$	gradient $=b$ , intercept $=\ln A$

To get the constant back, undo the $\ln$ . From a $\ln$ -graph, $k=e^{\text{intercept}}$ (and $A=e^{\text{intercept}}$ ). Stopping at $\ln k$ is only half an answer.

One quick way to tell which shape you're in

The exponential model, step by step

Worked example

$P=ae^{kt}$ : $\ln P$ against $t$ has gradient $\tfrac{1}{20}$ , cutting the axis at $(0,3)$

$P$ is a bacteria count at time $t$ hours. Find $k$ , and $a$ to 2 s.f.

Take $\ln$ : $\ln P=\ln a+kt$ , i.e. $\ln P=k\,t+\ln a$ .
Match to $Y=mX+c$ with $Y=\ln P$ , $X=t$ : gradient $=k$ , intercept $=\ln a$ .
So $k=\dfrac{1}{20}$ straight off the gradient.
Intercept gives $\ln a=3$ , so recover with $a=e^3\approx20$ .

Answer

k=\tfrac{1}{20},\quad a=e^3\approx20

(9709/33 Oct/Nov 2024 Q3(a))

Your turn— tap to reveal the worked answer (9709/33 Oct/Nov 2024 Q3(b))

With $P=e^3\,e^{t/20}$ , find how long it takes $P$ to double. Doubling means $e^{t/20}=2$ (the $e^3$ cancels). Take $\ln$ : $\dfrac{t}{20}=\ln 2$ .

Answer

t=20\ln 2\approx14\ \text{hours}

When you're given two points

Often you just get two points on the $\ln y$ -against- $x$ line.
The gradient between the two points is the constant. Then put one point back in to get the intercept.

Worked example

$ky=e^{cx}$ : line $\ln y$ vs $x$ through $(2.80,\,0.372)$ and $(5.10,\,2.21)$

Find $k$ and $c$ to 2 s.f.

Rearrange first: $ky=e^{cx}$ gives $\ln k+\ln y=cx$ , so $\ln y=cx-\ln k$ — gradient $c$ , intercept $-\ln k$ .
Gradient from the points: $c=\dfrac{2.21-0.372}{5.10-2.80}=0.80$ .
Sub a point for the intercept: $0.372=0.80(2.80)-\ln k\Rightarrow\ln k=1.87$ .
Recover: $k=e^{1.87}\approx6.5$ .

Answer

c=0.80,\quad k=6.5

(9709/33 May/Jun 2024 Q2)

Worked example

$y^2=Ae^{kx}$ : line $\ln y$ vs $x$ through $(1.5,\,1.2)$ and $(5.24,\,2.7)$

The hidden-power version. Find $A$ and $k$ to 2 d.p., and watch the $y^2$ .

Take $\ln$ : $\ln(y^2)=\ln A+kx$ , and $\ln(y^2)=2\ln y$ , so $2\ln y=kx+\ln A$ .
Divide by 2 to match the plotted axis $\ln y$ : $\ln y=\tfrac{k}{2}\,x+\tfrac12\ln A$ .
Gradient of the plotted line: $\dfrac{2.7-1.2}{5.24-1.5}=0.40$ , and that equals $\tfrac{k}{2}$ , so $k=0.80$ .
Intercept (sub a point): $1.2=0.40(1.5)+\tfrac12\ln A\Rightarrow\tfrac12\ln A=0.60$ , so $\ln A=1.20$ and $A=e^{1.2}\approx3.32$ .

Answer

A\approx3.32,\quad k=0.80

(9709/32 May/Jun 2020 Q2)

Read the axes — they tell you which constant is which

The biggest trap isn't the algebra. It is mis-reading what is plotted against what, so read the axis labels to see which constant is the gradient and which is the intercept.

Axes given	Means the model is	Watch for
$\ln y$ vs $x$	exponential $y=Ae^{bx}$	gradient is the rate
$\ln y$ vs $\ln x$	power $y=kx^n$	gradient is the power
$y$ vs $\ln x$	roles flipped	don't assume the standard shape

Common mistake

Three classic lost marks: (1) mixing up which constant is gradient and which is intercept; (2) forgetting to do

e^{\text{intercept}}

and leaving the answer as

\ln k

; (3) reading

y

\ln x

as if it were

\ln y

\ln x

, which swaps the roles completely.

Harder versions: when BOTH sides need a log first