Transforming Relationships to Linear Form

Exam code: 4037

Written by: Ashika|Reviewed by: Caroline Carroll|Updated 2 July 2026

Transforming Relationships in the Form y=ax^nvideo

Transforming Relationships in the Form y=ax^n

Transforming relationships in the form y=ax^n

How do I use logarithms to linearise a graph in the form y = axⁿ?

Logarithms can be used to linearise graphs of power functions
Suppose $y = a x^{n}$
- You can take logarithms of both sides
  - $\ln y = \ln (a x^{n})$
- You can split the right hand side into the sum of two logarithms
  - $\ln y = \ln a + \ln (x^{n})$
- You can bring down the power in the final term
  - $\ln y = \ln a + n \ln x$
$\ln y = \ln a + n \ln x$ is in linear form $Y = m X + c$
- $Y = \ln y$
- $X = \ln x$
- $m = n$
- $c = \ln a$

How can I use linearised form to find the unknown constants?

After linearising the function it will be in the form $\ln y = \ln a + n \ln x$
- $n$ is the gradient of the straight line graph
- $\ln a$ is the $y$ -intercept of the straight line graph
Once you know the value of the gradient of the straight line graph this is the value of $n$
You will need to find the value of $a$ by solving the equation $\ln a = c$

Worked example

The heights, $h$ metres, and the amount of time spent sleeping, $t$ hours, of a group of young giraffes can be modelled using $t = a h^{b}$ , where $a$ and $b$ are constants.

The graph of $\ln t$ against $\ln h$ is a straight line passing through the points (1, -0.9) and (4, -4.5).

Find the values of $a$ and $b$ , giving your answers in exact form.

Find the gradient of the straight line between the two coordinates.

$m = \frac{- 4.5 - - 0.9}{4 - 1} = - \frac{3.6}{3} = - 1.2$

Substitute $m = - 1.2$ , $x = \ln h$ and $y = \ln t$ into the equation of a straight line ( $y = m x + c$ ).

$\ln t = - 1.2 \ln h + c$

Substitute either coordinate in and rearrange to find $c$ .

$\begin{array}{rcl} - 0.9 & = & - 1.2 (1) + c \\ c & = & - 0.9 + 1.2 \\ c & = & 0.3 \end{array}$

Compare this to the linearised form of $t = a h^{b}$ . Take logarithms of both sides first and rearrange if you can't remember the correct form.

$\ln t = \ln (a h^{b}) \ln t = \ln a + b \ln h$

So $b$ is the value in front of $\ln h$ and $\ln a = 0.3$ . Solve $\ln a = 0.3$ by taking $e$ of both sides.

$\begin{array}{rcl} \ln a & = & 0.3 \\ e^{\ln a} & = & e^{0.3} \\ a & = & e^{0.3} \end{array}$

$a = e^{0.3}, b = - 1.2$

Transforming Relationships in the Form y=Ab^x

Transforming relationships in the form y=Ab^x

How do I use logarithms to linearise a graph in the form y = A(b^x)?

Logarithms can be used to linearise graphs of exponential functions
Suppose $y = A b^{x}$
- You can take logarithms of both sides
  - $\log y = \log (A b^{x})$
- You can split the right hand side into the sum of two logarithms
  - $\log y = \log A + \log (b^{x})$
- You can bring down the power in the final term
  - $\log y = \log A + x \log b$
$\log y = \log A + x \log b$ is in linear form $Y = m X + c$
- $Y = \log y$
- $X = x$
- $m = \log b$
- $c = \log A$

How can I use linearised form to find the unknown constants?

After linearising the function it will be in the form $\log y = \log A + x \log b$
- $\log b$ is the gradient of the straight line graph $(m)$
- $\log A$ is the $y$ -intercept of the straight line graph $(c)$
You will need to find the value of $A$ by solving the equation $\log A = c$
- The value of $c$ will either be given or will need to be found
You will need to find the value of $b$ by solving the equation $\log b = m$
- The value of $m$ will either be given or will need to be found

Worked example

Variables $x$ and $y$ are such that when $\lg y$ is plotted against $x$ , a straight line passing through the points $(2, 5)$ and $(5, 8)$ is obtained.

Show that $y = A \times b^{x}$ where $A$ and $b$ are constants to be found.

Find the gradient of the straight line between the two coordinates.

$m = \frac{8 - 5}{5 - 2} = \frac{3}{3} = 1$

Substitute $m = 1$ , $X = x$ and $Y = \lg y$ into the equation of a straight line ( $Y = m X + c$ ).

$\lg y = x + c$

Substitute either coordinate in and rearrange to find $c$ .

$\begin{array}{rcl} 5 & = & 2 + c \\ c & = & 3 \end{array}$

$\lg y = x + 3$

Linearise $y = A \times b^{x}$ .
Take logarithms of both sides first and rearrange if you can't remember the correct form. Take logarithms of both sides.

$\lg y = \lg (A b^{x})$

Split the right-hand side into the sum of two logarithms.

$\lg y = \lg A + \lg b^{x}$

Bring down the power in the final term.

$\lg y = \lg A + x \lg b$

Compare this to the equation of the line.

$\lg b = 1$ and $\lg A = 3$

Solve $\lg b = 1$ by raising 10 to the power of both sides.

$\begin{array}{rcl} \lg b & = & 1 \\ 10^{\lg b} & = & 10^{1} \\ b & = & 10 \end{array}$

Solve $\lg A = 3$ by raising 10 to the power of both sides.

$\begin{array}{rcl} \lg A & = & 3 \\ 10^{\lg A} & = & 10^{3} \\ A & = & 1000 \end{array}$

$y = 1000 \times 10^{x}$

Transforming Relationships in the Form y=ax^nvideo

Transforming relationships in the form y=ax^n

How do I use logarithms to linearise a graph in the form y = axn?

How can I use linearised form to find the unknown constants?

Transforming Relationships in the Form y=Ab^x

Transforming relationships in the form y=Ab^x

How do I use logarithms to linearise a graph in the form y = A(bx)?

How can I use linearised form to find the unknown constants?

Build on this topic

How do I use logarithms to linearise a graph in the form y = axⁿ?

How do I use logarithms to linearise a graph in the form y = A(b^x)?