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Summary

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:

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 $c = \log(a)$ .

Learn with Basics

Length:

Unit 1

Logarithms

Unit 2

Logarithms and non-linear data

Jump Ahead

Optional

Exponential models

Videos

Summary

Exercises

Select Lesson

Exam Board

Further kinematics

Application of forces

Projectiles

Forces and friction

Moments

Forces and motion

Variable acceleration

Constant acceleration

Modelling in mechanics

Normal distribution

Conditional probability

Correlation and hypothesis testing

Hypothesis testing I

Binomial distribution

Probability

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Measures of location and spread

Data collection

Vectors II

Numerical methods

Integration II

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Trigonometric equations

Trigonometry

Binomial expansion I

Circles

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Indices and surds

Proof I

Explainer Video

Summary

Exponential models

In a nutshell

The case where y=axny = ax^ny=axn​​

Example 1

The case where y=abxy = ab^xy=abx​​

Learn with Basics

Unit 1

Logarithms

Unit 2

Logarithms and non-linear data

Jump Ahead

Unit 3