Sec 4 · Additional Mathematics

Linear law

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1. Writing an equation as Y = mX + c

Equation	Straight-line form	$Y$	$X$	$m$	$c$
$y = ax^{2} + b$	$y = ax^{2} + b$	$y$	$x^{2}$	$a$	$b$
$y = ax^{2} + bx$	$\dfrac{y}{x} = ax + b$	$\dfrac{y}{x}$	$x$	$a$	$b$
$y = ax + \dfrac{b}{x}$	$xy = ax^{2} + b$	$xy$	$x^{2}$	$a$	$b$
$y = kx^{n}$	$\lg y = n\lg x + \lg k$	$\lg y$	$\lg x$	$n$	$\lg k$
$y = kb^{x}$	$\lg y = (\lg b)x + \lg k$	$\lg y$	$x$	$\lg b$	$\lg k$
$y = ae^{bx}$	$\ln y = bx + \ln a$	$\ln y$	$x$	$b$	$\ln a$

📖 Worked example

The variables $x$ and $y$ are related by the equation $y = ax + \dfrac{b}{x}$ , where $a$ and $b$ are constants. When $xy$ is plotted against $x^{2}$ , a straight line passing through the points $(3, 1)$ and $(7, 9)$ is obtained. Find the value of $b$ .

Step 1 of 5

Write the equation as Y = mX + c, with Y = xy and X = x²: what do you do?

Multiply both sides by x.

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2. Finding the constants from the line

y = kxⁿ: lg y against lg x

\displaystyle \lg y = n\lg x + \lg k

A straight line on axes lg x and lg y through (0.5, 0.35) and (1.5, −0.65), gradient −1, meeting the lg y axis at 0.85.

e.g.

\displaystyle n = \dfrac{-0.65 - 0.35}{1.5 - 0.5}

\displaystyle n = -1

intercept: $\lg k = 0.85$
$k = 10^{0.85}$ , which is $7.08$ (3 s.f.)

Read off	Undo the log
$\lg k = c$	$k = 10^{c}$
$\ln a = c$	$a = e^{c}$

📖 Worked example

The variables $x$ and $y$ are related by the equation $y = kx^{n}$ , where $k$ and $n$ are constants. When $\lg y$ is plotted against $\lg x$ , a straight line passing through the points $(0.5, 0.35)$ and $(1.5, -0.65)$ is obtained. Find the value of $k$ , correct to 3 significant figures.

Step 1 of 6

Take lg of both sides and use the laws of logarithms

\lg y = \lg k + n\lg x

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3. Exponential relationships

y = kbˣ: lg y against x, gradient 0.18

\displaystyle \lg b = 0.18

$b = 10^{0.18}$ , which is $1.51$ (3 s.f.)

y = aeᵇˣ: ln y against x, intercept 0.2

\displaystyle \ln a = 0.2

$a = e^{0.2}$ , which is $1.22$ (3 s.f.)

📖 Worked example

The variables $x$ and $y$ are related by the equation $y = ae^{bx}$ , where $a$ and $b$ are constants. When $\ln y$ is plotted against $x$ , a straight line passing through the points $(1, 1.4)$ and $(4, 5)$ is obtained. Find the value of $b$ .

Step 1 of 4

Take ln of both sides and use the laws of logarithms

\ln y = \ln a + bx

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Practice questions by skill

1. Writing an equation as Y = mX + c

Equation	Straight-line form	$Y$	$X$	$m$	$c$
$y = ax^{2} + b$	$y = ax^{2} + b$	$y$	$x^{2}$	$a$	$b$
$y = ax^{2} + bx$	$\dfrac{y}{x} = ax + b$	$\dfrac{y}{x}$	$x$	$a$	$b$
$y = ax + \dfrac{b}{x}$	$xy = ax^{2} + b$	$xy$	$x^{2}$	$a$	$b$
$y = kx^{n}$	$\lg y = n\lg x + \lg k$	$\lg y$	$\lg x$	$n$	$\lg k$
$y = kb^{x}$	$\lg y = (\lg b)x + \lg k$	$\lg y$	$x$	$\lg b$	$\lg k$
$y = ae^{bx}$	$\ln y = bx + \ln a$	$\ln y$	$x$	$b$	$\ln a$

📖 Worked example

Step 1 of 5

Write the equation as Y = mX + c, with Y = xy and X = x²: what do you do?

Multiply both sides by x.

Practise this →

2. Finding the constants from the line

y = kxⁿ: lg y against lg x

\displaystyle \lg y = n\lg x + \lg k

e.g.

\displaystyle n = \dfrac{-0.65 - 0.35}{1.5 - 0.5}

\displaystyle n = -1

intercept: $\lg k = 0.85$
$k = 10^{0.85}$ , which is $7.08$ (3 s.f.)

Read off	Undo the log
$\lg k = c$	$k = 10^{c}$
$\ln a = c$	$a = e^{c}$

📖 Worked example

Step 1 of 6

Take lg of both sides and use the laws of logarithms

\lg y = \lg k + n\lg x

Practise this →

3. Exponential relationships

y = kbˣ: lg y against x, gradient 0.18

\displaystyle \lg b = 0.18

$b = 10^{0.18}$ , which is $1.51$ (3 s.f.)

y = aeᵇˣ: ln y against x, intercept 0.2

\displaystyle \ln a = 0.2

$a = e^{0.2}$ , which is $1.22$ (3 s.f.)

📖 Worked example

Step 1 of 4

Take ln of both sides and use the laws of logarithms

\ln y = \ln a + bx

Practise this →