2 Log-linear graph paper

Ordinary graph paper has linear scales in both the horizontal $\left(x\right)$ and vertical $\left(y\right)$ directions. As we have seen, this can pose problems if the range of one of the variables, $y$ say, is very large. One way round this is to take the logarithm of the $y$ -values and re-plot on ordinary graph paper. Another common approach is to use log-linear graph paper in which the vertical scale is a non-linear logarithmic scale . Use of this special graph paper means that the original data can be plotted directly without the need to convert to logarithms which saves time and effort.

In log-linear graph paper the vertical axis is divided into a number of cycles . Each cycle corresponds to a jump in the data values by a factor of 10. For example, if the range of $y$ -values extends from (say) 1 to 100 (or equivalently $10^0$ to $10^2$ ) then 2-cycle log-linear paper would be required. If the $y$ -values extends from (say) 100 to 100,000 (or equivalently from $10^2$ to $10^5$ ) then 3-cycle log-linear paper would be used. Some other examples are given in Table 3:

Table 3

An example of 2-cycle log-linear graph paper is shown in Figure 14. We see that the horizontal scale is linear. The vertical scale is divided by lines denoted by 1,2,3, $\dots$ ,10,20,30, $\dots$ ,100. In the first cycle each of the horizontal blocks (separated by a slightly thicker line) is also divided according to a log-linear scale; so, for example, in the range $1\to 2$ we have 9 horizontal lines representing the values 1.1, 1.2, $\dots$ , 1.9. These subdivisions have been repeated (appropriately scaled) in blocks 2-3, 3-4, 4-5, 5-6, 6-7. The subdivisions have been omitted from blocks 7-8, 8-9, 9-10 for reasons of clarity. On this graph paper, we have noted the positions of $A:\left(1,2\right),B:\left(1,23\right),C:\left(4,23\right),D:\left(6,2.5\right),E:\left(3,61\right)$ .

Figure 14

Task!

On the 2-cycle log-linear graph paper (below) locate the positions of the points $F:\left(2,21\right)$ , $G:\left(2,51\right)$ , $H:\left(5,3.5\right)$ .

Answer

Example 12

It is thought that the relationship between two variables $x,y$ is exponential

$y=k{a}^x$

An experiment is performed and the following pairs of data values $\left(x,y\right)$ were obtained

Verify that the relation $y=k{a}^x$ is valid by plotting values on log-linear paper to obtain a set of points lying on a straight line. Estimate the values of $k,a$ .

Solution

First we rearrange the relation $y=k{a}^x$ by taking logarithms (to base 10).

$\therefore logy=log\left(k{a}^x\right)=logk+xloga$

So, if we define a new variable $Y\equiv logy$ then the relationship between $Y$ and $x$ will be linear $-$ its graph (on log-linear paper) should be a straight line. The vertical intercept of this line is $logk$ and the gradient of the line is $loga$ . Each of these can be obtained from the graph and the values of $a,k$ inferred.

When using log-linear graphs, the reader should keep in mind that, on the vertical axis, the values are not as written but the logarithms of those values.

We have plotted the points and drawn a straight line (as best we can) through them - see Figure 15. (We will see in a later Workbook ( HELM booklet  31) how we might improve on this subjective approach to fitting straight lines to data points). The line intersects the vertical axis at a value $log\left(3.13\right)$ and the gradient of the line is

$\frac{log96-log3.13}{5-0}=\frac{log\left(96∕3.13\right)}{5}=\frac{log30.67}{5}=0.297$

But the intercept is $logk$ so

$logk=log3.13\text{implying}k=3.13$

and the gradient is $loga$ so

$loga=0.297\text{implying}a=10^{0.297}=1.98$

We conclude that the relation between the $x,y$ variables is well modelled by the
relation $y=3.13{\left(1.98\right)}^x$ . If the points did not lie more-or-less on a straight line then we would conclude that the relationship was not of the form $y=k{a}^x$ .

Figure 15

Task!

Using a log-linear graph estimate the values of $k,a$ if it is assumed
that $y=k{a}^{-2x}$ and the data values connecting $x,y$ are:

First take logs of the relation $y=k{a}^{-2x}$ and introduce an appropriate new variable:

Answer

Now determine how many cycles are required in your log-linear paper:

Answer

Now plot the data values directly onto log-linear paper (supplied on the next page) and decide whether the relation $y=k{a}^{-2x}$ is acceptable:

Answer

Now, using knowledge of the intercept and the gradient, find the values of $k,a$ :

Answer

Use the log-linear graph sheets supplied on the following pages for these Exercises.

Exercises

1. Estimate the values of $k$ and $a$ if $y=k{a}^x$ represents the following set of data values:

   $x$	0.5	1	2	3	4
   $y$	5.93	8.8	19.36	42.59	93.70

2. Estimate the values of $k$ and $a$ if the relation $y=k{\left(a\right)}^{-x}$ is a good representation for the data values:

   $x$	2	2.5	3	3.5	4
   $y$	7.9	3.6	1.6	0.7	0.3

Answer