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 $$\log(a)$$​.

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:

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:

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

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 $$\log(a)$$​.