Logarithms and non-linear data

In a nutshell

Often you will be presented with sets of data which share a non-linear relationship. Logarithms can be used to identify and linearise non-linear trends in data.

Note: In most cases, you will see $\log(x)$ to mean $\log_{10}(x)$ .

The case where $y = ax^n$

Given a non-linear relationship of the form $y = ax^n$ , you can apply the logarithm with base $10$ (or indeed any base) and use logarithm laws to obtain a linear relationship between $\log(y)$ and $\log(x)$ :

$\begin{aligned}y &= ax^n\\\log(y) &= \log(ax^n)\\\log(y) &= n\log(x) + \log(a)\end{aligned}$

Comparing this with an equation of the form $Y = mX + c$ , you have that $\log(y)$ and $\log(x)$ are linearly related with $m=n$ and $c = \log(a)$ . The graph of $\log(y)$ against $\log(x)$ forms a straight line with slope $n$ and $y$ -intercept $\log(a)$ .

Maths; Exponentials and logarithms; KS5 Year 12; Logarithms and non-linear data

Example 1

The kinetic energy of a thrown baseball in joules is equal to $K = \dfrac12mv^2$ , where $m$ is the mass of the ball in $kg$ , and $v$ is the speed of the ball in $ms^{-1}$ . Throughout a sequence of experiments, the speed and kinetic energy of the ball are measured to $1\space d.p.$ at various times:

$K$	$39.1$	$44.4$	$50.4$	$92.3$	$123.4$
$v$	$22.9$	$24.4$	$26.0$	$35.2$	$40.7$

Create a table of values for $\log(K)$ against $\log(v)$ to $2\space d.p.$ , and use it to plot $\log(K)$ against $\log(v)$ . Draw a line of best fit through these points, and use it to estimate the value of $m$ to $2\space d.p.$ .

Your table of values will look like:

$\log(K)$	$1.59$	$1.65$	$1.70$	$1.96$	$2.09$
$\log(v)$	$1.36$	$1.39$	$1.41$	$1.54$	$1.61$

Graphing these points, and putting a line of best fit through them (which should have a slope of approximately $2$ ) looks like:

Maths; Exponentials and logarithms; KS5 Year 12; Logarithms and non-linear data

By taking the logarithm of both sides of $K=\frac12mv^2$ , you have

$\begin{aligned}\log(K)&=\log\bigg(\frac12mv^2\bigg)\\\log(K)&=\log\bigg(\frac12m\bigg)+\log(v^2)\\\log(K)&=\log\bigg(\frac12m\bigg)+2\log(v)\\\log(K)&=2\log(v)+\log\bigg(\frac12m\bigg)\end{aligned}$

This gives the linear equation for the line above. The $y$ -intercept is approximately $-1.13$ . Thus you find that $-1.13 = \log\left(\dfrac12m\right)$ , and so $m = 2 \times 10^{-1.13} = 0.148\dots$

Therefore, to $2\space d.p.$ the mass of the baseball is approximately $\underline{0.15kg}$ .

The case where $y = ab^x$

Given a non-linear relationship of the form $y = ab^x$ , you can apply the logarithm with base $10$ (or indeed any base) and use logarithm laws to obtain a linear relationship between $\log(y)$ and $x$ :

$\begin{aligned}y &= ab^x\\\log(y) &= \log(ab^x)\\\log(y) &= x\log(b) + \log(a)\end{aligned}$

Comparing this with an equation of the form $Y = mX + c$ , you have that $\log(y)$ and $x$ are linearly related with $m = \log(b)$ and $c = \log(a)$ . So the slope of the graph of the equation $\log(y) = \log(b)x + \log(a)$ is $\log(b)$ , and the $y$ -intercept is $\log(a)$ .

Maths; Exponentials and logarithms; KS5 Year 12; Logarithms and non-linear data