7 Logarithmic relationships

Experimental psychology is concerned with observing and measuring human response to various stimuli. In particular, sensations of light, colour, sound, taste, touch and muscular tension are produced when an external stimulus acts on the associated sense. A nineteenth century German, Ernst Weber, conducted experiments involving sensations of heat, light and sound and associated stimuli. Weber measured the response of subjects, in a laboratory setting, to input stimuli measured in terms of energy or some other physical attribute and discovered that:

1. No sensation is felt until the stimulus reaches a certain value, known as the threshold value.
2. After this threshold is reached an increase in stimulus produces an increase in sensation.
3. This increase in sensation occurs at a diminishing rate as the stimulus is increased.
1. Do Weber’s results suggest a linear or non-linear relationship between sensation and stimulus? Sketch a graph of sensation against stimulus according to Weber’s results.
2. Consider whether an exponential function or a growth-to-a-limit function might be an appropriate model.
1. Non-linearity is required by observation (3).

2. An exponential-type of growth is not appropriate for a model consistent with these experimental results, since we need a diminishing rate of growth in sensation as the stimulus increases. A growth-to-a-limit type of function is not appropriate since the data, at least over the range of Weber’s experiments, do not suggest that there is a limit to the sensation with continuing increase in stimulus; only that the increase in sensation occurs more and more slowly.

A late nineteenth century German scientist, Gustav Fechner, studied Weber’s results. Fechner suggested that an appropriate function modelling Weber’s findings would be logarithmic. He suggested that the variation in sensation $\left(S\right)$ with the stimulus input $\left(P\right)$ is modelled by

$\phantom{\rule{2em}{0ex}}S=Alog\left(P∕T\right)\phantom{\rule{2em}{0ex}}\left(0

where $T$ represents the threshold of stimulus input below which there is no sensation and $A$ is a constant. Note that when $P=T,log\left(P∕T\right)=log\left(1\right)=0$ , so this function is consistent with item 1. of Weber’s results. Recall also that $log$ means logarithm to base 10, so when $P=10T,S=Alog\left(10\right)=A$ . When P = 100T, $S=Alog\left(100\right)=2A$ . The logarithmic function predicts that a tenfold increase in the stimulus input from $T$ to $10T$ will result in the same change in sensation as a further tenfold increase in stimulus input to $100T$ . Each tenfold change is stimulus results in a doubling of sensation. So, although sensation is predicted to increase with stimulus, the stimulus has to increase at a faster and faster rate (i.e. exponentially) to achieve a given change in sensation. These points are consistent with items 2. and 3. of Weber’s findings. Fechner’s suggestion, that the logarithmic function is an appropriate one for a model of the relationship between sensation and stimulus, seems reasonable. Note that the logarithmic function suggested by Weber is not defined for zero stimulus but we are only interested in the model at and above the threshold stimulus, i.e. for values of the logarithm equal to and above zero. Note also that the logarithmic function is useful for looking at changes in sensation relative to stimulus values other than the threshold stimulus. According to Rule 2a in Table 2 on page 42, Fechner’s sensation function may be written

$\phantom{\rule{2em}{0ex}}S=Alog\left(P∕T\right)=A\left[log\left(P\right)-log\left(T\right)\right]\phantom{\rule{2em}{0ex}}\left(P\ge T>0\right).$

Suppose that the sensation has the value ${S}_{1}$ at ${P}_{1}$ and ${S}_{2}$ at ${P}_{2}$ , so that

$\phantom{\rule{2em}{0ex}}{S}_{1}=A\left[log\left({P}_{1}\right)-log\left(T\right)\right]\phantom{\rule{2em}{0ex}}\left({P}_{1}\ge T>0\right),$

and

$\phantom{\rule{2em}{0ex}}{S}_{2}=A\left[log\left({P}_{2}\right)-log\left(T\right)\right]\phantom{\rule{2em}{0ex}}\left({P}_{2}\ge T>0\right).$

If we subtract the first of these two equations from the second, we get

$\phantom{\rule{2em}{0ex}}{S}_{2}-{S}_{1}=A\left[log\left({P}_{2}\right)-log\left({P}_{1}\right)\right]=Alog\left({P}_{2}∕{P}_{1}\right),$

where Rule 2a of Table 2 has been used again for the last step. According to this form of equation, the change in sensation between two stimuli values depends on the ratio of the stimuli values.

$\phantom{\rule{2em}{0ex}}S=Alog\left(P∕T\right)\phantom{\rule{2em}{0ex}}\left(1\ge T>0\right).$

Divide both sides by A:

$\phantom{\rule{2em}{0ex}}\frac{S}{A}=log\frac{P}{T}\phantom{\rule{2em}{0ex}}\left(1\ge T>0\right).$

‘Undo’ the logarithm on both sides by raising 10 to the power of each side:

$\phantom{\rule{2em}{0ex}}1{0}^{S∕A}=1{0}^{log\left(P∕T\right)}=\frac{P}{T}\phantom{\rule{2em}{0ex}}\left(1\ge T>0\right),$  using Rule 4b of Table 2.

So $P=T×1{0}^{S∕A}\phantom{\rule{2em}{0ex}}\left(1\ge T>0\right)$ which is an exponential relationship between stimulus and sensation.

A logarithmic relationship between sensation and stimulus therefore implies an exponential relationship between stimulus and sensation. The relationship may be written in two different forms with the variables playing opposite roles in the two functions.

The logarithmic relationship between sensation and stimulus is known as the Weber-Fechner Law of Sensation. The idea that a mathematical function could describe our sensations was startling when first propounded. Indeed it may seem quite amazing to you now. Moreover it doesn’t always work. Nevertheless the idea has been quite fruitful. Out of it has come much quantitative experimental psychology of interest to sound engineers. For example, it relates to the sensation of the loudness of sound. Sound level is expressed on a logarithmic scale. At a frequency of 1 kHz an increase of 10 dB corresponds to a doubling of loudness.

Given a relationship between $y$ and $x$ of the form $y=3log\left(\frac{x}{4}\right)\phantom{\rule{2em}{0ex}}\left(x\ge 4\right),$ find the relationship between $x$ and $y$ .

One way of answering is to compare with the example preceding this task. We have $y$ in place of $S$ , $x$ in place of $P$ , 3 in place of $A$ , 4 in place of $T$ . So it is possible to write down immediately

$\phantom{\rule{2em}{0ex}}x=4×1{0}^{y∕3}\phantom{\rule{2em}{0ex}}\left(y\ge 0\right)$

Alternatively we can manipulate the given expression algebraically.

Starting with $\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}y=3log\left(x∕4\right),$ divide both sides by 3 to give $\phantom{\rule{1em}{0ex}}y∕3=log\left(x∕4\right)$ .

Raise 10 to the power of each side to eliminate the log, so that $\phantom{\rule{1em}{0ex}}1{0}^{y∕3}=x∕4$ .

Multiply both sides by 4 and rearrange, to obtain $x=4×1{0}^{y∕3}$ , as before.

The associated range is the result of the fact that $x\ge 4$ , so $1{0}^{y∕3}\ge 1$ , so $y∕3>0$ which means $y>0$ .