Not in G2 · G3 Additional Maths only

Chapter 8 · Additional Maths · G3 Linear law

Linear law: turning curves into straight lines

A straight line is the easiest graph to read. Linear law shows you how to rewrite a curved relationship as Y = mX + c, so that two points or one plotted line are enough to find the unknown constants. By the end of this chapter you can choose the right axes, work with logarithms when the model is a power or an exponential, and estimate values from the line.

Your outcome tracker0 / 6

Let's go! Tick an outcome once you can pass its exam check. Saved on this device.

Chapter
8

Figure 1 · a power law drawn as a straight line

1 outcomes · 1 page · every step shown ✦

Everything is Y = mX + c

Rewrite the relation so that one expression is Y and another is X. The gradient and the intercept must be constants: they may not contain $x$ or $y$ .

Logs straighten powers

$y = ax^n$ becomes $\lg y = n\lg x + \lg a$ and $y = kb^x$ becomes $\lg y = (\lg b)x + \lg k$ . Read $a = 10^c$ and $b = 10^m$ from the line.

Convert before you read

A value given for $y$ on a $\lg y$ axis is $\lg$ of that value. Convert it to the plotted variable, read the line, then convert back.

By the end of this chapter

Rewrite a relation such as y = ax^2 + b, y = a/x + b or y = ax + b/x in the form Y = mX + c and name X, Y, the gradient and the intercept.
Use lg or ln to straighten a power model y = ax^n or an exponential model y = kb^x, and recover the constants with 10^c and 10^m.
Find the constants from two points of the line, algebraically (paper 1 style) or from a plotted line (paper 2 style).
Convert a target value to the plotted variable before reading the line, and spot and correct one wrongly recorded data point.
Use a second line drawn on the same graph to solve a related equation, and apply the method to real data.

✶

Start here

Maths In Real Life

Two situations where measured data follow a curve. The pictures come first; the straight-line graph in each solution comes from the same numbers.

$Side view of a school laboratory bench. A straight wooden ramp rises from the lower left to the upper right, a small trolley sits on the lower part of the ramp and a motion sensor is fixed at the top end of the ramp. The distance d from the sensor to the trolley is marked along the ramp$

d

Maths In Real Life 1

A trolley on a ramp

A trolley that speeds up steadily on a ramp covers a distance that grows with the square of the time, so plotting the distance against the time squared gives a straight line whose intercept is the starting distance.

A motion sensor at the top of a ramp measures the distance $d$ centimetres from the sensor to a trolley. The trolley is released and rolls down the ramp. At $t$ seconds after release, $d = pt^{2} + q$ , where $p$ and $q$ are constants. The sensor records $d = 25$ when $t = 1$ and $d = 65$ when $t = 3$ .

(a)A graph of $d$ against $t^{2}$ is a straight line through the two readings. Find the gradient and the intercept of the line.
(b)Hence find the values of $p$ and $q$ .
(c)The sensor stops recording when the trolley is $100$ centimetres from it. Use the straight line to find the time at which this happens.

Show full solutionHide solution

Colour keyplotted value X or Y worked outgradient mintercept cfinal constants

(a)

Straight-line shape: plot

d

against

t^{2}

\displaystyle X_1\;= 1^{2} = \textcolor{#010103}{1}

The first coordinate plotted for the point with

t = 1

\displaystyle Y_1\;= \textcolor{#010103}{25}

\displaystyle X_2\;= 3^{2} = \textcolor{#010103}{9}

Same for the second point.

\displaystyle Y_2\;= \textcolor{#010103}{65}

\displaystyle m\;= \dfrac{Y_2 - Y_1}{X_2 - X_1} = \dfrac{65 - 25}{9 - 1} = \textcolor{#010104}{5}

The gradient of the straight line through the two plotted points.

\displaystyle c\;= Y_1 - mX_1 = 25 - 5 = \textcolor{#010105}{20}

Put one point back into

Y = mX + c

to find the intercept.

(b)

Straight-line shape: plot

d

against

t^{2}

\displaystyle X_1\;= 1^{2} = \textcolor{#010103}{1}

The first coordinate plotted for the point with

t = 1

\displaystyle Y_1\;= \textcolor{#010103}{25}

\displaystyle X_2\;= 3^{2} = \textcolor{#010103}{9}

Same for the second point.

\displaystyle Y_2\;= \textcolor{#010103}{65}

\displaystyle m\;= \dfrac{Y_2 - Y_1}{X_2 - X_1} = \dfrac{65 - 25}{9 - 1} = \textcolor{#010104}{5}

The gradient of the straight line through the two plotted points.

\displaystyle c\;= Y_1 - mX_1 = 25 - 5 = \textcolor{#010105}{20}

Put one point back into

Y = mX + c

to find the intercept.

The model has

p = m

and

q = c

, so

p = 5

and

q = 20

(c)

\displaystyle d\;= 5t^{2} + 20

The straight line, with

m

and

c

read from the model.

\displaystyle d\;= 100

Here the plotted

Y

is just

d

, so put in the given value.

\displaystyle 100 = 5t^{2} + 20

Put the plotted

Y

-value into the line equation.

\displaystyle t^{2}\;= \dfrac{100 - 20}{5} = \textcolor{#010105}{16}

Solve the line equation for the plotted

X

\displaystyle t\;= \sqrt{16} = \textcolor{#010106}{4}

Since

X = t^{2}

and

t

is positive, take the positive square root.

Answer

(a)

m = 5

c = 20

(b)

p = 5

q = 20

(c)

t = 4

$A top view of a round glass petri dish on a laboratory bench. Small round creamy colonies of bacteria are scattered over the pale agar, and the number of colonies is counted in the study.$

N = pq^{t}

Maths In Real Life 2

Bacteria growing in a dish

Bacteria in a dish multiply by splitting, so the count grows by the same factor every hour. A graph of the count itself bends sharply upward, but its logarithm rises in a straight line, which is why scientists plot logs of counts.

A scientist counts the bacteria $N$ in a petri dish $t$ hours after a sample is placed in it. The counts follow $N = pq^t$ , where $p$ and $q$ are constants. After $1$ hour the count is $120$ and after $3$ hours it is $1080$ .

(a)By plotting $\lg N$ against $t$ , find the values of $p$ and $q$ .
(b)Use the line to estimate the time when the count reaches $3240$ .

Show full solutionHide solution

Colour keygradient mintercept cfinal constants

(a)

Straight-line shape: plot

\lg N

against

t

For the point with

t = 1

and

N = 120

, the plotted

X

-value is

1

and the plotted

Y

-value is

\lg 120

Each data point becomes a point

(X, Y)

on the

\lg

scale.

For the point with

t = 3

and

N = 1080

, the plotted values are

3

and

\lg 1080

\displaystyle m\;= \dfrac{Y_2 - Y_1}{X_2 - X_1} = \dfrac{\lg 1080 - \lg 120}{3 - 1} = \dfrac{\lg 9}{2}

Use

\lg a - \lg b = \lg \dfrac{a}{b}

on the top.

\displaystyle \lg 9 = \lg 3^{2} = 2 \lg 3

Write

9

3^{2}

and bring the power down.

\displaystyle m\;= \dfrac{2 \lg 3}{2} = \textcolor{#010104}{\lg 3}

The common factor cancels.

\displaystyle c\;= Y_1 - mX_1 = \lg 120 - \lg 3 = \textcolor{#010105}{\lg 40}

Put one point back into

Y = mX + c

, then use

\lg a - \lg b = \lg \dfrac{a}{b}

The gradient is

m = \lg q

and the intercept is

c = \lg p

\displaystyle \lg q\;= \lg 3

\displaystyle \lg p\;= \lg 40

q = 3

and

p = 40

(b)

\displaystyle \lg N\;= \left(\lg 3\right)t + \lg 40

The straight line, with

m

and

c

read from the model.

\displaystyle \lg N\;= \lg 3240

Convert the given value to the plotted variable before using the line.

\displaystyle \lg 3240 = \left(\lg 3\right)t + \lg 40

Put the plotted

Y

-value into the line equation.

\displaystyle t\;= \dfrac{\lg 3240 - \lg 40}{\lg 3} = \dfrac{\lg 81}{\lg 3}

Rearrange for the unknown, then use

\lg a - \lg b = \lg \dfrac{a}{b}

\displaystyle \lg 81 = \lg 3^{4} = 4 \lg 3

Write

81

3^{4}

and bring the power down.

\displaystyle t\;= \dfrac{4 \lg 3}{\lg 3} = \textcolor{#010106}{4}

The

\lg

factors cancel.

Answer

(a) the two points

(1, \lg 120)

(3, \lg 1080)

;

m = \lg 3

c = \lg 40

;

p = 40

q = 3

(b)

t = 4

Supports K341-G2-05

Turning a relation into a straight line

A curve is hard to read and a straight line is easy. Linear law is the skill of rewriting a curved relation as a straight line, so that two points, or one drawn line, are enough to find the unknown constants.

Revision. A straight line

y = mx + c

has gradient

m

, the rise divided by the run, and cuts the vertical axis at

c

. In this chapter the axes are called

X

and

Y

because they may carry whole expressions, such as

x^{2}

\dfrac{1}{x}

, but the ideas are exactly the same.

Links To

The gradient and intercept of a straight line were met in Chapter 7: Coordinate geometry: lines, areas and circles

The Core Idea

Reduce to $Y = mX + c$ .

Rearrange the relation so that one expression is $Y$ and another is $X$ , with the gradient $m$ and the intercept $c$ being constants: numbers that contain neither $x$ nor $y$ . Then $Y$ plotted against $X$ is a straight line. Read off $m$ and $c$ , then use them to recover the constants of the original relation.

The usual moves are to multiply or divide both sides by $x$ , to treat a power such as $x^{2}$ or a reciprocal such as $\dfrac{1}{x}$ as one variable, or (next section) to take logarithms. After the move, the right-hand side must be a number times $X$ plus a number.

Example 1.1

Adding a constant to x squared

The relation between $x$ and $y$ is $y = 3x^{2} + 2$ . State the quantities that should be plotted on the vertical and the horizontal axes to obtain a straight line, and find the gradient and the vertical intercept of that line.

Show solutionHide solution

Colour keygradient mintercept c

\displaystyle y = 3x^{2} + 2

The relation we are given.

The right-hand side is already a straight-line shape if we treat

x^{2}

as one variable.

Compare with

Y = mX + c

. So plot

y

on the vertical axis against

x^{2}

on the horizontal axis.

\displaystyle y\;= \textcolor{#010104}{3x^{2}} \textcolor{#010105}{+ 2}

Same shape as

Y = mX + c

: the number multiplying the variable is

m

, the number added is

c

So the gradient is

m = 3

and the vertical intercept is

c = 2

Answer

Plot

y

against

x^{2}

m = 3

c = 2

Your Turn 1.1

The relation between $x$ and $y$ is $y = 4x^{2} - 5$ . State the quantities that should be plotted on the vertical and the horizontal axes to obtain a straight line, and find the gradient and the vertical intercept of that line.

Show full solutionHide solution

Colour keygradient mintercept c

\displaystyle y = 4x^{2} - 5

The relation we are given.

The right-hand side is already a straight-line shape if we treat

x^{2}

as one variable.

Compare with

Y = mX + c

. So plot

y

on the vertical axis against

x^{2}

on the horizontal axis.

\displaystyle y\;= \textcolor{#010104}{4x^{2}} \textcolor{#010105}{-5}

Same shape as

Y = mX + c

: the number multiplying the variable is

m

, the number added is

c

So the gradient is

m = 4

and the vertical intercept is

c = -5

Answer

Plot

y

against

x^{2}

m = 4

c = -5

Example 1.2

A reciprocal as the variable

A student wants to test the relation $y = \dfrac{6}{x} + 1$ with a straight-line graph. Say which quantity goes on each axis, and find the gradient and the vertical intercept of the line.

Show solutionHide solution

Colour keygradient mintercept c

\displaystyle y = \dfrac{6}{x} + 1

The relation we are given.

Write

\dfrac{6}{x}

6 \times \dfrac{1}{x}

, so

\dfrac{1}{x}

plays the part of the variable.

Compare with

Y = mX + c

. So plot

y

on the vertical axis against

\dfrac{1}{x}

on the horizontal axis.

\displaystyle y\;= \textcolor{#010104}{6\dfrac{1}{x}} \textcolor{#010105}{+ 1}

Same shape as

Y = mX + c

: the number multiplying the variable is

m

, the number added is

c

So the gradient is

m = 6

and the vertical intercept is

c = 1

Answer

Plot

y

against

\dfrac{1}{x}

m = 6

c = 1

Your Turn 1.2

A student wants to test the relation $y = \dfrac{8}{x} - 3$ with a straight-line graph. Say which quantity goes on each axis, and find the gradient and the vertical intercept of the line.

Show full solutionHide solution

Colour keygradient mintercept c

\displaystyle y = \dfrac{8}{x} - 3

The relation we are given.

Write

\dfrac{8}{x}

8 \times \dfrac{1}{x}

, so

\dfrac{1}{x}

plays the part of the variable.

Compare with

Y = mX + c

. So plot

y

on the vertical axis against

\dfrac{1}{x}

on the horizontal axis.

\displaystyle y\;= \textcolor{#010104}{8\dfrac{1}{x}} \textcolor{#010105}{-3}

Same shape as

Y = mX + c

: the number multiplying the variable is

m

, the number added is

c

So the gradient is

m = 8

and the vertical intercept is

c = -3

Answer

Plot

y

against

\dfrac{1}{x}

m = 8

c = -3

SGSchoolKaki Tip

Look for the part of the relation that is multiplied by a number: that part is $X$ (or $Y$ ). The number that is added or subtracted on its own is the intercept $c$ .

Example 1.3

Clearing a fraction first

The variables $x$ and $y$ satisfy $y = 2x + \dfrac{3}{x}$ . By choosing suitable expressions for $X$ and $Y$ , reduce the relation to a straight line. State $X$ and $Y$ , and find the gradient and the vertical intercept.

Show solutionHide solution

Colour keygradient mintercept c

\displaystyle y = 2x + \dfrac{3}{x}

The relation we are given.

\displaystyle xy = x\left(2x + \dfrac{3}{x}\right) = 2x^{2} + 3

Multiply both sides by

x

to clear the fraction in

\dfrac{3}{x}

Compare with

Y = mX + c

. So plot

xy

on the vertical axis against

x^{2}

on the horizontal axis.

\displaystyle xy\;= \textcolor{#010104}{2x^{2}} \textcolor{#010105}{+ 3}

Same shape as

Y = mX + c

: the number multiplying the variable is

m

, the number added is

c

So the gradient is

m = 2

and the vertical intercept is

c = 3

Answer

Plot

xy

against

x^{2}

m = 2

c = 3

Your Turn 1.3

The variables $x$ and $y$ satisfy $y = 5x - \dfrac{4}{x}$ . By choosing suitable expressions for $X$ and $Y$ , reduce the relation to a straight line. State $X$ and $Y$ , and find the gradient and the vertical intercept.

Show full solutionHide solution

Colour keygradient mintercept c

\displaystyle y = 5x - \dfrac{4}{x}

The relation we are given.

\displaystyle xy = x\left(5x - \dfrac{4}{x}\right) = 5x^{2} - 4

Multiply both sides by

x

to clear the fraction in

\dfrac{4}{x}

Compare with

Y = mX + c

. So plot

xy

on the vertical axis against

x^{2}

on the horizontal axis.

\displaystyle xy\;= \textcolor{#010104}{5x^{2}} \textcolor{#010105}{-4}

Same shape as

Y = mX + c

: the number multiplying the variable is

m

, the number added is

c

So the gradient is

m = 5

and the vertical intercept is

c = -4

Answer

Plot

xy

against

x^{2}

m = 5

c = -4

The gradient and the intercept must be constants. The relation

y = ax + \dfrac{b}{x}

does not give a straight line when

y

is plotted against

x

, because

\dfrac{b}{x}

changes as

x

changes. Multiply both sides by

x

first: then

xy = ax^{2} + b

is a straight line when

xy

is plotted against

x^{2}

Example 1.4

Dividing by x

The relation $y = 3x^{2} - 2x$ is to be drawn as a straight line. Say which quantity goes on each axis (one of them needs a division), and find the gradient and the vertical intercept of the line.

Show solutionHide solution

Colour keygradient mintercept c

\displaystyle y = 3x^{2} - 2x

The relation we are given.

\displaystyle \dfrac{y}{x} = 3x - 2

Divide both sides by

x

so that the right-hand side is a straight-line shape in

x

Compare with

Y = mX + c

. So plot

\dfrac{y}{x}

on the vertical axis against

x

on the horizontal axis.

\displaystyle \dfrac{y}{x}\;= \textcolor{#010104}{3x} \textcolor{#010105}{-2}

Same shape as

Y = mX + c

: the number multiplying the variable is

m

, the number added is

c

So the gradient is

m = 3

and the vertical intercept is

c = -2

Answer

Plot

\dfrac{y}{x}

against

x

m = 3

c = -2

Your Turn 1.4

The relation $y = 2x^{2} + 5x$ is to be drawn as a straight line. Say which quantity goes on each axis (one of them needs a division), and find the gradient and the vertical intercept of the line.

Show full solutionHide solution

Colour keygradient mintercept c

\displaystyle y = 2x^{2} + 5x

The relation we are given.

\displaystyle \dfrac{y}{x} = 2x + 5

Divide both sides by

x

so that the right-hand side is a straight-line shape in

x

Compare with

Y = mX + c

. So plot

\dfrac{y}{x}

on the vertical axis against

x

on the horizontal axis.

\displaystyle \dfrac{y}{x}\;= \textcolor{#010104}{2x} \textcolor{#010105}{+ 5}

Same shape as

Y = mX + c

: the number multiplying the variable is

m

, the number added is

c

So the gradient is

m = 2

and the vertical intercept is

c = 5

Answer

Plot

\dfrac{y}{x}

against

x

m = 2

c = 5

Think About It

Why is

y = 4x + \dfrac{3}{x}

not a straight line when

y

is plotted against

x

? Test it with three values of

x

Show the full workingHide the working

Work out

y

at three equally spaced values of

x

1

2

and

3

. Equal steps in

x

should give equal steps in

y

on a straight line.

A straight line rises by the same amount for every equal step.

\displaystyle 4 \times 1 + \dfrac{3}{1} = 7

The value of

y

when

x = 1

\displaystyle 4 \times 2 + \dfrac{3}{2} = 9.5

The value of

y

when

x = 2

\displaystyle 4 \times 3 + \dfrac{3}{3} = 13

The value of

y

when

x = 3

\displaystyle 9.5 - 7 = 2.5

The first step up.

\displaystyle 13 - 9.5 = 3.5

The second step up.

Answer

The steps

2.5

and

3.5

are different, so the points do not lie on a straight line: the term

\dfrac{3}{x}

changes the gradient.

Which axes straighten which relation
Relation	Plot Y against X	Gradient	Intercept	Graph
$y = ax^{2} + b$	$y$ against $x^{2}$	$a$	$b$
$y = \dfrac{a}{x} + b$	$y$ against $\dfrac{1}{x}$	$a$	$b$
$y = a\sqrt{x} + b$	$y$ against $\sqrt{x}$	$a$	$b$
$y = ax + \dfrac{b}{x}$	$xy$ against $x^{2}$	$a$	$b$	multiply by $x$ first
$y = ax^{2} + bx$	$\dfrac{y}{x}$ against $x$	$a$	$b$	divide by $x$ first
$y = ax^{n}$	$\lg y$ against $\lg x$	$n$	$\lg a$
$y = kb^{x}$	$\lg y$ against $x$	$\lg b$	$\lg k$

Example 1.5

A square root as the variable

The variables are related by $y = 4\sqrt{x} - 1$ . Which quantity should be plotted against which to give a straight line? State the gradient and the vertical intercept.

Show solutionHide solution

Colour keygradient mintercept c

\displaystyle y = 4\sqrt{x} - 1

The relation we are given.

The variable on the right is

\sqrt{x}

, to the power 1.

Compare with

Y = mX + c

. So plot

y

on the vertical axis against

\sqrt{x}

on the horizontal axis.

\displaystyle y\;= \textcolor{#010104}{4\sqrt{x}} \textcolor{#010105}{-1}

Same shape as

Y = mX + c

: the number multiplying the variable is

m

, the number added is

c

So the gradient is

m = 4

and the vertical intercept is

c = -1

Answer

Plot

y

against

\sqrt{x}

m = 4

c = -1

Your Turn 1.5

The variables are related by $y = 6\sqrt{x} - 2$ . Which quantity should be plotted against which to give a straight line? State the gradient and the vertical intercept.

Show full solutionHide solution

Colour keygradient mintercept c

\displaystyle y = 6\sqrt{x} - 2

The relation we are given.

The variable on the right is

\sqrt{x}

, to the power 1.

Compare with

Y = mX + c

. So plot

y

on the vertical axis against

\sqrt{x}

on the horizontal axis.

\displaystyle y\;= \textcolor{#010104}{6\sqrt{x}} \textcolor{#010105}{-2}

Same shape as

Y = mX + c

: the number multiplying the variable is

m

, the number added is

c

So the gradient is

m = 6

and the vertical intercept is

c = -2

Answer

Plot

y

against

\sqrt{x}

m = 6

c = -2

Example 1.6

A cube plus a constant

A scientist records $x$ and $y$ and believes that $y = 2x^{3} - 3$ . Describe a straight-line graph that tests this belief, and find its gradient and intercept.

Show solutionHide solution

Colour keygradient mintercept c

\displaystyle y = 2x^{3} - 3

The relation we are given.

The right-hand side is already a straight-line shape if we treat

x^{3}

as one variable.

Compare with

Y = mX + c

. So plot

y

on the vertical axis against

x^{3}

on the horizontal axis.

\displaystyle y\;= \textcolor{#010104}{2x^{3}} \textcolor{#010105}{-3}

Same shape as

Y = mX + c

: the number multiplying the variable is

m

, the number added is

c

So the gradient is

m = 2

and the vertical intercept is

c = -3

Answer

Plot

y

against

x^{3}

m = 2

c = -3

Your Turn 1.6

A scientist records $x$ and $y$ and believes that $y = 5x^{3} + 4$ . Describe a straight-line graph that tests this belief, and find its gradient and intercept.

Show full solutionHide solution

Colour keygradient mintercept c

\displaystyle y = 5x^{3} + 4

The relation we are given.

The right-hand side is already a straight-line shape if we treat

x^{3}

as one variable.

Compare with

Y = mX + c

. So plot

y

on the vertical axis against

x^{3}

on the horizontal axis.

\displaystyle y\;= \textcolor{#010104}{5x^{3}} \textcolor{#010105}{+ 4}

Same shape as

Y = mX + c

: the number multiplying the variable is

m

, the number added is

c

So the gradient is

m = 5

and the vertical intercept is

c = 4

Answer

Plot

y

against

x^{3}

m = 5

c = 4

Example 1.7

One relation, two straight lines

The relation $y = \dfrac{6}{x} + 5$ can be reduced to a straight line in two different ways. First, put $y$ on the vertical axis: state what goes on the horizontal axis, and the gradient and the intercept. Second, put $xy$ on the vertical axis: state what goes on the horizontal axis, and the gradient and the intercept.

(i)Put $y$ on the vertical axis. State what goes on the horizontal axis, and the gradient and the intercept.
(ii)Put $xy$ on the vertical axis. State what goes on the horizontal axis, and the gradient and the intercept.

Show solutionHide solution

Colour keygradient mintercept c

(i)

\displaystyle y = \dfrac{6}{x} + 5

The relation we are given.

Write

\dfrac{6}{x}

6 \times \dfrac{1}{x}

, so

\dfrac{1}{x}

plays the part of the variable.

Compare with

Y = mX + c

. So plot

y

on the vertical axis against

\dfrac{1}{x}

on the horizontal axis.

\displaystyle y\;= \textcolor{#010104}{6\dfrac{1}{x}} \textcolor{#010105}{+ 5}

Same shape as

Y = mX + c

: the number multiplying the variable is

m

, the number added is

c

So the gradient is

m = 6

and the vertical intercept is

c = 5

(ii)

\displaystyle xy\;= x\left(\dfrac{6}{x} + 5\right) = 6 + 5x

Multiply both sides by

x

to clear the fraction: the left-hand side becomes

xy

\displaystyle xy\;= \textcolor{#010104}{5x} \textcolor{#010105}{ + 6}

Put the

x

term first: the number multiplying

x

m

, the number added is

c

So plot

xy

on the vertical axis against

x

on the horizontal axis: the gradient is

m = 5

and the intercept is

c = 6

Answer

(i) Plot

y

against

\dfrac{1}{x}

m = 6

c = 5

(ii) Plot

xy

against

x

m = 5

c = 6

Your Turn 1.7

The relation $y = \dfrac{9}{x} - 2$ can be reduced to a straight line in two different ways. First, put $y$ on the vertical axis: state what goes on the horizontal axis, and the gradient and the intercept. Second, put $xy$ on the vertical axis: state what goes on the horizontal axis, and the gradient and the intercept.

(i)Put $y$ on the vertical axis. State what goes on the horizontal axis, and the gradient and the intercept.
(ii)Put $xy$ on the vertical axis. State what goes on the horizontal axis, and the gradient and the intercept.

Show full solutionHide solution

Colour keygradient mintercept c

(i)

\displaystyle y = \dfrac{9}{x} - 2

The relation we are given.

Write

\dfrac{9}{x}

9 \times \dfrac{1}{x}

, so

\dfrac{1}{x}

plays the part of the variable.

Compare with

Y = mX + c

. So plot

y

on the vertical axis against

\dfrac{1}{x}

on the horizontal axis.

\displaystyle y\;= \textcolor{#010104}{9\dfrac{1}{x}} \textcolor{#010105}{-2}

Same shape as

Y = mX + c

: the number multiplying the variable is

m

, the number added is

c

So the gradient is

m = 9

and the vertical intercept is

c = -2

(ii)

\displaystyle xy\;= x\left(\dfrac{9}{x} - 2\right) = 9 - 2x

Multiply both sides by

x

to clear the fraction: the left-hand side becomes

xy

\displaystyle xy\;= \textcolor{#010104}{-2x} \textcolor{#010105}{ + 9}

Put the

x

term first: the number multiplying

x

m

, the number added is

c

So plot

xy

on the vertical axis against

x

on the horizontal axis: the gradient is

m = -2

and the intercept is

c = 9

Answer

(i) Plot

y

against

\dfrac{1}{x}

m = 9

c = -2

(ii) Plot

xy

against

x

m = -2

c = 9

Supports K341-G2-05

Power and exponential models with logarithms

When $x$ is a power ( $y = ax^{n}$ ) or the variable is in an exponent ( $y = kb^{x}$ ), no rearrangement with $\times$ or $\div$ gives a straight line. Logarithms do: they turn products into sums and powers into multiples.

Revision. For positive numbers,

\lg (ab) = \lg a + \lg b

\lg \dfrac{a}{b} = \lg a - \lg b

and

\lg a^{n} = n \lg a

. The same laws hold for

\ln

, where

\ln e = 1

. Also

10^{\lg a} = a

, which undoes a logarithm.

Links To

The laws of logarithms were developed in Chapter 6: Exponential and logarithmic functions

The Core Idea

Take $\lg$ of both sides, then match $Y = mX + c$ .

Power model: $y = ax^{n}$ gives $\lg y = n \lg x + \lg a$ . Plot $\lg y$ against $\lg x$ : gradient $n$ , intercept $\lg a$ , so $a = 10^{c}$ . Exponential model: $y = kb^{x}$ gives $\lg y = (\lg b)x + \lg k$ . Plot $\lg y$ against $x$ : gradient $\lg b$ , intercept $\lg k$ , so $b = 10^{m}$ and $k = 10^{c}$ . Both need $y > 0$ , with $\textcolor{#010101}{a > 0}$ (power) or $k > 0$ (exponential); the power model also needs $x > 0$ , because $\lg x$ must exist.

SGSchoolKaki Tip

Say which logarithm you are using. With $\lg$ (base $10$ ) the constants come back with $10^{c}$ and $10^{m}$ ; with $\ln$ they come back with $e^{c}$ and $e^{m}$ . Never write the intercept $c$ as if it were $a$ : $c = \lg a$ .

Example 2.1

A power model

The relation between $x$ and $y$ is $y = 3x^{2}$ . Use logarithms ( $\lg$ ) to reduce the relation to a straight line. State what is plotted on each axis, and the gradient and the vertical intercept.

Show solutionHide solution

Colour keygradient mintercept c

\displaystyle y = 3x^{2}

The relation we are given.

\displaystyle \lg y = \lg\left(3x^{2}\right) = \lg 3 + 2\lg x

Take

\lg

of both sides, then use the product law and the power law.

Compare with

Y = mX + c

. So plot

\lg y

on the vertical axis against

\lg x

on the horizontal axis.

\displaystyle \lg y\;= \textcolor{#010104}{2\lg x} \textcolor{#010105}{+ \lg 3}

Same shape as

Y = mX + c

: the number multiplying the variable is

m

, the number added is

c

So the gradient is

2

and the vertical intercept is

\lg 3

Answer

Plot

\lg y

against

\lg x

m = 2

c = \lg 3

Your Turn 2.1

The relation between $x$ and $y$ is $y = 6x^{5}$ . Use logarithms ( $\lg$ ) to reduce the relation to a straight line. State what is plotted on each axis, and the gradient and the vertical intercept.

Show full solutionHide solution

Colour keygradient mintercept c

\displaystyle y = 6x^{5}

The relation we are given.

\displaystyle \lg y = \lg\left(6x^{5}\right) = \lg 6 + 5\lg x

Take

\lg

of both sides, then use the product law and the power law.

Compare with

Y = mX + c

. So plot

\lg y

on the vertical axis against

\lg x

on the horizontal axis.

\displaystyle \lg y\;= \textcolor{#010104}{5\lg x} \textcolor{#010105}{+ \lg 6}

Same shape as

Y = mX + c

: the number multiplying the variable is

m

, the number added is

c

So the gradient is

5

and the vertical intercept is

\lg 6

Answer

Plot

\lg y

against

\lg x

m = 5

c = \lg 6

Example 2.2

An exponential model

The relation between $x$ and $y$ is $y = 2 \times 5^{x}$ . Use logarithms ( $\lg$ ) to reduce the relation to a straight line. State what is plotted on each axis, and the gradient and the vertical intercept.

Show solutionHide solution

Colour keygradient mintercept c

\displaystyle y = 2 \times 5^{x}

The relation we are given.

\displaystyle \lg y = \lg\left(2 \times 5^{x}\right) = \lg 2 + x\left(\lg 5\right)

Take

\lg

of both sides, then the product law and the power law (the power is

x

Compare with

Y = mX + c

. So plot

\lg y

on the vertical axis against

x

on the horizontal axis.

\displaystyle \lg y\;= \textcolor{#010104}{\left(\lg 5\right)x} \textcolor{#010105}{+ \lg 2}

Same shape as

Y = mX + c

: the number multiplying the variable is

m

, the number added is

c

So the gradient is

\lg 5

and the vertical intercept is

\lg 2

Answer

Plot

\lg y

against

x

m = \lg 5

c = \lg 2

Your Turn 2.2

The relation between $x$ and $y$ is $y = 3 \times 4^{x}$ . Use logarithms ( $\lg$ ) to reduce the relation to a straight line. State what is plotted on each axis, and the gradient and the vertical intercept.

Show full solutionHide solution

Colour keygradient mintercept c

\displaystyle y = 3 \times 4^{x}

The relation we are given.

\displaystyle \lg y = \lg\left(3 \times 4^{x}\right) = \lg 3 + x\left(\lg 4\right)

Take

\lg

of both sides, then the product law and the power law (the power is

x

Compare with

Y = mX + c

. So plot

\lg y

on the vertical axis against

x

on the horizontal axis.

\displaystyle \lg y\;= \textcolor{#010104}{\left(\lg 4\right)x} \textcolor{#010105}{+ \lg 3}

Same shape as

Y = mX + c

: the number multiplying the variable is

m

, the number added is

c

So the gradient is

\lg 4

and the vertical intercept is

\lg 3

Answer

Plot

\lg y

against

x

m = \lg 4

c = \lg 3

Example 2.3

The natural exponential

The relation between $x$ and $y$ is $y = 4e^{3x}$ . Use natural logarithms to reduce the relation to a straight line. State what is plotted on each axis, and the gradient and the vertical intercept.

Show solutionHide solution

Colour keygradient mintercept c

\displaystyle y = 4e^{3x}

The relation we are given.

\displaystyle \ln y = \ln\left(4e^{3x}\right) = \ln 4 + 3x

Take

\ln

of both sides;

\ln e = 1

removes the

e

Compare with

Y = mX + c

. So plot

\ln y

on the vertical axis against

x

on the horizontal axis.

\displaystyle \ln y\;= \textcolor{#010104}{3x} \textcolor{#010105}{+ \ln 4}

Same shape as

Y = mX + c

: the number multiplying the variable is

m

, the number added is

c

So the gradient is

3

and the vertical intercept is

\ln 4

Answer

Plot

\ln y

against

x

m = 3

c = \ln 4

Your Turn 2.3

The relation between $x$ and $y$ is $y = 2e^{5x}$ . Use natural logarithms to reduce the relation to a straight line. State what is plotted on each axis, and the gradient and the vertical intercept.

Show full solutionHide solution

Colour keygradient mintercept c

\displaystyle y = 2e^{5x}

The relation we are given.

\displaystyle \ln y = \ln\left(2e^{5x}\right) = \ln 2 + 5x

Take

\ln

of both sides;

\ln e = 1

removes the

e

Compare with

Y = mX + c

. So plot

\ln y

on the vertical axis against

x

on the horizontal axis.

\displaystyle \ln y\;= \textcolor{#010104}{5x} \textcolor{#010105}{+ \ln 2}

Same shape as

Y = mX + c

: the number multiplying the variable is

m

, the number added is

c

So the gradient is

5

and the vertical intercept is

\ln 2

Answer

Plot

\ln y

against

x

m = 5

c = \ln 2

The four logarithm models
Model	Plot Y against X	Gradient	Intercept
$y = ax^{n}$	$\lg y$ against $\lg x$	$n$	$\lg a$
$y = kb^{x}$	$\lg y$ against $x$	$\lg b$	$\lg k$
$y = ae^{kx}$	$\ln y$ against $x$	$k$	$\ln a$
$y = ax^{n} + r$	$\lg (y - r)$ against $\lg x$	$n$	$\lg a$

Think About It

In the model

y = ax^{n} + r

, why must

r

be taken away before taking logarithms? Test

\lg (10 + 90)

Show the full workingHide the working

Compare

\lg (a + b)

with

\lg a + \lg b

for

a = 10

and

b = 90

The log laws turn products into sums, not sums into anything.

\displaystyle \lg (10 + 90)\;= \lg 100 = 2

Add first, then take the logarithm.

\displaystyle \lg 10 + \lg 90\;= \lg 900 = \lg 100 + \lg 9 = 2 + \lg 9

Adding logarithms multiplies the numbers.

Answer

Since

\lg 9 \neq 0

the two results differ, so

\lg (a + b)

is not

\lg a + \lg b

. Subtract

r

first, so that one product is left to take the logarithm of.

Example 2.4

A shifted power model

The relation between $x$ and $y$ is $y = 2x^{3} + 5$ . Reduce the relation to a straight line using logarithms ( $\lg$ ). State what is plotted on each axis, and the gradient and the vertical intercept.

Show solutionHide solution

Colour keygradient mintercept c

\displaystyle y = 2x^{3} + 5

The relation we are given.

\displaystyle y - 5 = 2x^{3}

Move the constant to the left so that only the power term is on the right.

\displaystyle \lg\left(y - 5\right) = \lg 2 + 3\lg x

Now take

\lg

of both sides.

Compare with

Y = mX + c

. So plot

\lg \left(y -5\right)

on the vertical axis against

\lg x

on the horizontal axis.

\displaystyle \lg \left(y -5\right)\;= \textcolor{#010104}{3\lg x} \textcolor{#010105}{+ \lg 2}

Same shape as

Y = mX + c

: the number multiplying the variable is

m

, the number added is

c

So the gradient is

3

and the vertical intercept is

\lg 2

Answer

Plot

\lg \left(y -5\right)

against

\lg x

m = 3

c = \lg 2

Your Turn 2.4

The relation between $x$ and $y$ is $y = 3x^{2} - 4$ . Reduce the relation to a straight line using logarithms ( $\lg$ ). State what is plotted on each axis, and the gradient and the vertical intercept.

Show full solutionHide solution

Colour keygradient mintercept c

\displaystyle y = 3x^{2} - 4

The relation we are given.

\displaystyle y + 4 = 3x^{2}

Move the constant to the left so that only the power term is on the right.

\displaystyle \lg\left(y + 4\right) = \lg 3 + 2\lg x

Now take

\lg

of both sides.

Compare with

Y = mX + c

. So plot

\lg \left(y +4\right)

on the vertical axis against

\lg x

on the horizontal axis.

\displaystyle \lg \left(y +4\right)\;= \textcolor{#010104}{2\lg x} \textcolor{#010105}{+ \lg 3}

Same shape as

Y = mX + c

: the number multiplying the variable is

m

, the number added is

c

So the gradient is

2

and the vertical intercept is

\lg 3

Answer

Plot

\lg \left(y +4\right)

against

\lg x

m = 2

c = \lg 3

Example 2.5

A negative power

The intensity $y$ of a signal at distance $x$ follows $y = 5x^{-2}$ . Reduce the relation to a straight line using logarithms ( $\lg$ ), then state the axes, the gradient and the vertical intercept.

Show solutionHide solution

Colour keygradient mintercept c

\displaystyle y = 5x^{-2}

The relation we are given.

\displaystyle \lg y = \lg\left(5x^{-2}\right) = \lg 5 - 2\lg x

Take

\lg

of both sides, then use the product law and the power law.

Compare with

Y = mX + c

. So plot

\lg y

on the vertical axis against

\lg x

on the horizontal axis.

\displaystyle \lg y\;= \textcolor{#010104}{-2\lg x} \textcolor{#010105}{+ \lg 5}

Same shape as

Y = mX + c

: the number multiplying the variable is

m

, the number added is

c

So the gradient is

-2

and the vertical intercept is

\lg 5

Answer

Plot

\lg y

against

\lg x

m = -2

c = \lg 5

Your Turn 2.5

The intensity $y$ of a signal at distance $x$ follows $y = 2x^{-3}$ . Reduce the relation to a straight line using logarithms ( $\lg$ ), then state the axes, the gradient and the vertical intercept.

Show full solutionHide solution

Colour keygradient mintercept c

\displaystyle y = 2x^{-3}

The relation we are given.

\displaystyle \lg y = \lg\left(2x^{-3}\right) = \lg 2 - 3\lg x

Take

\lg

of both sides, then use the product law and the power law.

Compare with

Y = mX + c

. So plot

\lg y

on the vertical axis against

\lg x

on the horizontal axis.

\displaystyle \lg y\;= \textcolor{#010104}{-3\lg x} \textcolor{#010105}{+ \lg 2}

Same shape as

Y = mX + c

: the number multiplying the variable is

m

, the number added is

c

So the gradient is

-3

and the vertical intercept is

\lg 2

Answer

Plot

\lg y

against

\lg x

m = -3

c = \lg 2

Example 2.6

Which axes make the data straight?

Four readings of $y$ are recorded at four values of $x$ : $x = 1$ gives $y = 6$ , $x = 2$ gives $y = 12$ , $x = 3$ gives $y = 24$ and $x = 4$ gives $y = 48$ . The readings follow either $y = ab^{x}$ or $y = ax^{n}$ , where $a$ , $b$ and $n$ are constants. By examining how the readings change, decide whether $\lg y$ should be plotted against $x$ or against $\lg x$ to give a straight line.

Show solutionHide solution

Colour keyplotted value X or Y worked out

Divide each reading of

y

by the one before it.

\displaystyle \dfrac{y_2}{y_1}\;= \dfrac{12}{6} = \textcolor{#010103}{2}

\displaystyle \dfrac{y_3}{y_2}\;= \dfrac{24}{12} = \textcolor{#010103}{2}

\displaystyle \dfrac{y_4}{y_3}\;= \dfrac{48}{24} = \textcolor{#010103}{2}

The

x

-values go up in equal steps of

1

and

y

is multiplied by

2

each time. Then

\lg y

goes up by the same amount each time:

\displaystyle \lg y_2 - \lg y_1\;= \lg 12 - \lg 6 = \lg 2

\displaystyle \lg y_3 - \lg y_2\;= \lg 24 - \lg 12 = \lg 2

\displaystyle \lg y_4 - \lg y_3\;= \lg 48 - \lg 24 = \lg 2

Equal steps in

x

give equal steps in

\lg y

, so

\lg y

plotted against

x

is a straight line.

Answer

Plot

\lg y

against

x

Your Turn 2.6

Four readings of $y$ are recorded at four values of $x$ : $x = 1$ gives $y = 6$ , $x = 2$ gives $y = 48$ , $x = 4$ gives $y = 384$ and $x = 8$ gives $y = 3072$ . The readings follow either $y = ab^{x}$ or $y = ax^{n}$ , where $a$ , $b$ and $n$ are constants. By examining how the readings change, decide whether $\lg y$ should be plotted against $x$ or against $\lg x$ to give a straight line.

Show full solutionHide solution

Colour keyplotted value X or Y worked out

Divide each reading of

y

by the one before it.

\displaystyle \dfrac{y_2}{y_1}\;= \dfrac{48}{6} = \textcolor{#010103}{8}

\displaystyle \dfrac{y_3}{y_2}\;= \dfrac{384}{48} = \textcolor{#010103}{8}

\displaystyle \dfrac{y_4}{y_3}\;= \dfrac{3072}{384} = \textcolor{#010103}{8}

Now divide each

x

-value by the one before it.

\displaystyle \dfrac{x_2}{x_1}\;= \dfrac{2}{1} = 2

\displaystyle \dfrac{x_3}{x_2}\;= \dfrac{4}{2} = 2

\displaystyle \dfrac{x_4}{x_3}\;= \dfrac{8}{4} = 2

Each

x

-value is multiplied by

2

and each

y

-value by

8

. Taking logarithms turns every multiplication into an addition, so both

\lg x

and

\lg y

rise by equal steps:

\displaystyle \lg x_2 - \lg x_1\;= \lg 2 - \lg 1 = \lg 2

\displaystyle \lg x_3 - \lg x_2\;= \lg 4 - \lg 2 = \lg 2

\displaystyle \lg x_4 - \lg x_3\;= \lg 8 - \lg 4 = \lg 2

\displaystyle \lg y_2 - \lg y_1\;= \lg 48 - \lg 6 = \lg 8

\displaystyle \lg y_3 - \lg y_2\;= \lg 384 - \lg 48 = \lg 8

\displaystyle \lg y_4 - \lg y_3\;= \lg 3072 - \lg 384 = \lg 8

Equal steps in

\lg x

give equal steps in

\lg y

, so

\lg y

plotted against

\lg x

is a straight line.

Answer

Plot

\lg y

against

\lg x

Locked · Chapter 8

You've read the free 24%. Unlock the rest of Chapter 8.

Still to read in this chapter:

Two points
Plot and read
Errors and second line
In context
Formula panel
Exercise

$9 one-time · all 17 chapters · no subscription

Worked examples, formulas and exam checks.
The whole A-Math syllabus.

ONE PRICE17chapters

Premium members get all 17 chapters included.

Notes version 1.0 · Last updated 3 Nov 2026 · Chapter 8 of 17