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. 

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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 $$c = \log(a)$$​.  

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 $$Y=mX+c$$.​

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 $$2\space d.p.$$:

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 $$x$$:

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

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

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