2 Linearisation of exponential functions

This subsection relates to the description of log-linear plots covered in Section 6.6.

Frequently in engineering, the question arises of how the parameters of an exponential function might be found from given data. The method follows from the fact that it is possible to ‘undo’ the exponential function and obtain a linear function by means of the logarithmic function. Before showing the implications of this method, it may be necessary to remind you of some rules for manipulating logarithms and exponentials. These are summarised in Table 1 on the next page, which exactly matches the general list provided in Key Point 8 in Section 6.3 (page 22.) Table 1:  Rules for manipulating base $\text{e}$ logarithms and exponentials

 Number Rule Number Rule 1a $ln\left(xy\right)=ln\left(x\right)+ln\left(y\right)$ 1b ${\text{e}}^{x}×{\text{e}}^{y}={\text{e}}^{x+y}$ 2a $ln\left(x∕y\right)=ln\left(x\right)-ln\left(y\right)$ 2b ${\text{e}}^{x}/{\text{e}}^{y}={\text{e}}^{x-y}$ 3a $ln\left({x}^{y}\right)=yln\left(x\right)$ 3b ${\left({\text{e}}^{x}\right)}^{y}={\text{e}}^{xy}$ 4a $ln\left({\text{e}}^{x}\right)=x$ 4b ${\text{e}}^{ln\left(x\right)}=x$ 5a $ln\left(\text{e}\right)=1$ 5b ${\text{e}}^{1}=\text{e}$ 6a $ln\left(1\right)=0$ 6b ${\text{e}}^{0}=1$

We will try ‘undoing’ the exponential in the particular example

$\phantom{\rule{2em}{0ex}}P=12{e}^{0.1t}$

We take the natural logarithm ( $ln$ ) of both sides, which means logarithm to the base $e$ . So

$\phantom{\rule{2em}{0ex}}ln\left(P\right)=ln\left(12{e}^{0.1t}\right)$

The result of using Rule 1a in Table 1 is

$\phantom{\rule{2em}{0ex}}ln\left(P\right)=ln\left(12\right)+ln\left({e}^{0.1t}\right).$

The natural logarithmic functions ‘undoes’ the exponential function, so by Rule 4a,

$\phantom{\rule{2em}{0ex}}ln\left({e}^{0.1t}\right)=0.1t$

and the original equation for $P$ becomes

$\phantom{\rule{2em}{0ex}}ln\left(P\right)=ln\left(12\right)+0.1t.$

Compare this with the general form of a linear function $y=ax+b$ .

$\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}y\phantom{\rule{1em}{0ex}}=\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}ax\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\phantom{\rule{1em}{0ex}}b$

$\phantom{\rule{1em}{0ex}}\phantom{\rule{2em}{0ex}}↓\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}↓\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}↓$

$\phantom{\rule{2em}{0ex}}ln\left(P\right)=0.1t+ln\left(12\right)$

If we regard $ln\left(P\right)$ as equivalent to $y$ , 0.1 as equivalent to the constant $a,t$ as equivalent to $x$ , and $ln\left(12\right)$ as equivalent to the constant $b$ , then we can identify a linear relationship between $ln\left(P\right)$ and $t$ . A plot of $ln\left(P\right)$ against $t$ should result in a straight line, of slope 0.1, which crosses the $ln\left(P\right)$ axis at $ln\left(12\right)$ . (Such a plot is called a log-linear or log-lin plot.) This is not particularly interesting here because we know the values 12 and 0.1 already.

Suppose, though, we want to try using the general form of the exponential function

$\phantom{\rule{2em}{0ex}}P=a{e}^{bt}\phantom{\rule{2em}{0ex}}\left(c\le t\le d\right)$

to create a continuous model for a population for which we have some discrete data. The first thing to do is to take logarithms of both sides

$\phantom{\rule{2em}{0ex}}ln\left(P\right)=ln\left(a{e}^{bt}\right)\phantom{\rule{2em}{0ex}}\left(c\le t\le d\right).$

Rule 1 from Table 1 then gives

$\phantom{\rule{2em}{0ex}}ln\left(P\right)=ln\left(a\right)+ln\left({e}^{bt}\right)\phantom{\rule{2em}{0ex}}\left(c\le t\le d\right).$

But, by Rule 4a, $ln\left({e}^{bt}\right)=bt$ , so this means that

$\phantom{\rule{2em}{0ex}}ln\left(P\right)=ln\left(a\right)+bt\phantom{\rule{2em}{0ex}}\left(c\le t\le d\right).$

So, given some ‘population versus time’ data, for which you believe can be modelled by some version of the exponential function, plot the natural logarithm of population against time. If the exponential function is appropriate, the resulting data points should lie on or near a straight line. The slope of the straight line will give an estimate for $b$ and the intercept with the $ln\left(P\right)$ axis will give an estimate for $ln\left(a\right).$ You will have carried out a logarithmic transformation of the original data for $P$ . We say the original variation has been linearised .

A similar procedure will work also if any exponential function rather than the base $\text{e}$ exponential function is used. For example, suppose that we try to use the function

$\phantom{\rule{2em}{0ex}}P=A×{2}^{Bt}\phantom{\rule{2em}{0ex}}\left(C\le t\le D\right),$

where A and B are constant parameters to be derived from the given data. We can take natural logarithms again to give

$\phantom{\rule{2em}{0ex}}ln\left(P\right)=ln\left(A×{2}^{Bt}\right)\phantom{\rule{2em}{0ex}}\left(C\le t\le D\right).$

Rule 1a from Table 1 then gives

$\phantom{\rule{2em}{0ex}}ln\left(P\right)=ln\left(A\right)+ln\left({2}^{Bt}\right)\phantom{\rule{2em}{0ex}}\left(C\le t\le D\right).$

Rule 3a then gives

$\phantom{\rule{2em}{0ex}}ln\left({2}^{Bt}\right)=Btln\left(2\right)=Bln\left(2\right)\phantom{\rule{1em}{0ex}}t$

and so

$\phantom{\rule{2em}{0ex}}ln\left(P\right)=ln\left(A\right)+Bln\left(2\right)\phantom{\rule{1em}{0ex}}t\phantom{\rule{2em}{0ex}}\left(C\le t\le D\right).$

Again we have a straight line graph with the same intercept as before, $lnA$ , but this time with slope $Bln\left(2\right)$ .

The amount of money $£M$ to which $£1$ grows after earning interest of 5 $%$ p.a. for $N$ years is worked out as

$\phantom{\rule{2em}{0ex}}M=1.0{5}^{N}$

Find a linearised form of this equation.

Take natural logarithms of both sides.

$\phantom{\rule{2em}{0ex}}ln\left(M\right)=ln\left(1.0{5}^{N}\right).$

Rule 3b gives

$\phantom{\rule{2em}{0ex}}ln\left(M\right)=Nln\left(1.05\right).$

So a plot of $ln\left(M\right)$ against $N$ would be a straight line passing through $\left(0,0\right)$ with slope $ln\left(1.05\right).$

The linearisation procedure also works if logarithms other than natural logarithms are used. We start again with

$\phantom{\rule{2em}{0ex}}P=A×{2}^{Bt}\phantom{\rule{2em}{0ex}}\left(C\le t\le D\right).$

and will take logarithms to base 10 instead of natural logarithms. Table 2 presents the laws of logarithms and indices (based on Key Point 8 page 22) interpreted for ${log}_{10}$ .

Table 2 : Rules for manipulating base 10 logarithms and exponentials

 Number Rule Number Rule 1a $\phantom{\rule{1em}{0ex}}{log}_{10}\left(AB\right)={log}_{10}A+{log}_{10}B$ 1b $\phantom{\rule{1em}{0ex}}1{0}^{A}1{0}^{B}=1{0}^{A+B}$ 2a $\phantom{\rule{1em}{0ex}}{log}_{10}\left(A∕B\right)={log}_{10}A-{log}_{10}B$ 2b $\phantom{\rule{1em}{0ex}}1{0}^{A}∕1{0}^{B}=1{0}^{A-B}$ 3a $\phantom{\rule{1em}{0ex}}{log}_{10}\left({A}^{k}\right)=k{log}_{10}A$ 3b $\phantom{\rule{1em}{0ex}}{\left(1{0}^{A}\right)}^{k}=1{0}^{kA}$ 4a $\phantom{\rule{1em}{0ex}}{log}_{10}\left(1{0}^{A}\right)=A$ 4b $\phantom{\rule{1em}{0ex}}1{0}^{{log}_{10}A}=A$ 5a $\phantom{\rule{1em}{0ex}}{log}_{10}10=1$ 5b $\phantom{\rule{1em}{0ex}}1{0}^{1}=10\phantom{\rule{2em}{0ex}}$ 6a $\phantom{\rule{1em}{0ex}}{log}_{10}1=0$ 6b $\phantom{\rule{1em}{0ex}}1{0}^{0}=1$

Taking logs of $P=A×{2}^{Bt}$ gives:

$\phantom{\rule{2em}{0ex}}{log}_{10}\left(P\right)={log}_{10}\left(A×{2}^{Bt}\right)\phantom{\rule{2em}{0ex}}\left(C\le t\le D\right).$

Rule 1a from Table 2 then gives

$\phantom{\rule{2em}{0ex}}{log}_{10}\left(P\right)={log}_{10}\left(A\right)+{log}_{10}\left({2}^{Bt}\right)\phantom{\rule{2em}{0ex}}\left(C\le t\le D\right).$

Use of Rule 3a gives the result

$\phantom{\rule{2em}{0ex}}{log}_{10}\left(P\right)={log}_{10}\left(A\right)+B{log}_{10}\left(2\right)\phantom{\rule{1em}{0ex}}t\phantom{\rule{2em}{0ex}}\left(C\le t\le D\right).$

1. Write down the straight line function corresponding to taking logarithms of the general exponential function

$\phantom{\rule{2em}{0ex}}P=a{e}^{bt}\phantom{\rule{2em}{0ex}}\left(c\le t\le d\right)$

by taking logarithms to base 10.

2. Write down the slope of this line.
1. ${log}_{10}\left(P\right)={log}_{10}\left(a\right)+\left(b{log}_{10}\left(e\right)\right)t\phantom{\rule{2em}{0ex}}\left(c\le t\le d\right)$
2. $b{log}_{10}\left(e\right)$

It is not usually necessary to declare the subscript 10 when indicating logarithms to base 10. If you meet the term ‘log’ it will probably imply “to the base 10”. In the remainder of this Section, the subscript 10 is dropped where ${log}_{10}$ is implied.