Exponential models

In a nutshell

Experimental data often forms polynomial and exponential models in the form $y=ax^n$ and $y=ab^x$ . It is not practical to examine these non-linear trends. Logarithms can be used to code regression lines which practically define trends and allow for a more precise form of extrapolation.

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 any base) and use logarithm laws to obtain a linear relationship between $\log(y)$ and $\log(x)$ :

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

$\boxed{\log(y) = n\log(x) + \log(a)}$

This equation is a linear relationship in the form $Y = mX + c$ , where $Y=\log(y)$ , $X = \log(x)$ , $m=n$ and $c = \log(a)$ .

Maths; Correlation and hypothesis testing; KS5 Year 13; Exponential models

Example 1

The kinetic energy of a ball, denoted as $K$ , is given by the formula $K = \dfrac12mv^2$ where $m$ is the mass $(kg)$ , and $v$ is the velocity $(ms^{-1})$ . An experiment is undertaken to measure the kinetic energy of the ball as its velocity increases with the results given in the table below to $1$ decimal place. Using this data, find the mass of the ball to $2$ decimal places. Do this by coding a regression line in the form $Y=mX+c$ .

$K (J)$	$39.1$	$44.4$	$50.4$	$92.3$	$123.4$
$v \ (ms^{-1})$	$22.9$	$24.4$	$26.0$	$35.2$	$40.7$

The relationship between kinetic energy and velocity is in the form:

$\begin{aligned} K &= \left(\dfrac12 m\right)v^2 \\\\ y& = (a)x^n \end{aligned}$ 

Code this relationship to fit a regression line:

$\begin{aligned} \log(y) &= n\log(x)+\log(a) \\ \log(K) &= 2\log(v) +\log\left(\dfrac12m\right) \\ Y&= mX + c \end{aligned}$ 

Create a table of $\log(K)$ against $\log(v)$ to $2\space d.p.$ :

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

Plot the graph of $\log(K) = 2\log(v) +\log(\dfrac12m)$ and draw a line of best fit. From this graph, extrapolate the $y$ -intercept:

Maths; Correlation and hypothesis testing; KS5 Year 13; Exponential models

Compare the equation of the line to $Y=mX+c$ to identify the $y$ -intercept:

$\begin{aligned}Y&=mX + c \\ \log(K)&=2\log(v) + \log\bigg(\frac12m\bigg) \\ c &= \log\bigg(\dfrac12m\bigg)\end{aligned}$ 

Calculate the mass, $m$ :

$\begin{aligned} \log\left(\dfrac12m\right)&=-1.13\\ \\ \dfrac12m &= 10^{-1.13}\\\\ m &= 2 \times 10^{-1.13} = 0.148\dots \end{aligned}$

$\space \ \underline{m= 0.15kg \ (2 \ d.p.)}$ 

Note: $\log(x)$ is equal to $\log_{10}(x)$ .

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 $x$ :

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

$\boxed{\log(y) = x\log(b) + \log(a)}$

This forms a linear relationship of $Y = mX + c$ where $Y=\log(y)$ , $X=x$ , $m = \log(b)$ and $c = \log(a)$ .

Maths; Correlation and hypothesis testing; KS5 Year 13; Exponential models