Logarithm scales are often used when there is a large range of values involved with the variables under consideration. Here is a simple example to motivate the idea of a log scale:
Suppose we consider the set of five ordered pairs shown here:
$x$x | $1$1 | $2$2 | $3$3 | $4$4 | $5$5 |
---|---|---|---|---|---|
$y$y | $20$20 | $200$200 | $3631$3631 | $52481$52481 | $250000$250000 |
A plot of the five points would be difficult to manage because of the range of the $y$y values.
See the following graph plot, the scale on the y-axis is so huge, that we lost a lot of the information from the first 3 points.
One way forward would be to develop a strategy that enables the reader to access the information indirectly. For example, we could plot the base $10$10 logarithm of $y$y against $x$x, with values shown in a new table.
$x$x | $1$1 | $2$2 | $3$3 | $4$4 | $5$5 |
---|---|---|---|---|---|
$\log_{10}y$log10y | $1.301$1.301 | $2.301$2.301 | $3.560$3.560 | $4.720$4.720 | $5.398$5.398 |
Even though the $y$y values are far more manageable in this form, we need to remember that the actual data points are those in the first table. That is to say the actual $y$y values, correct to $3$3 decimal places at least, are given by $10^{1.301},10^{2.301},10^{3.560},10^{4.720}$101.301,102.301,103.560,104.720 and $10^{5.398}$105.398.
Using the logarithm of the $y$y values gives us the following graph.
But of course, we can only retrieve the original data values by using a formula.
The idea that scientists and others struck upon was to leave the numbers alone (keep the $y$y values as they originally were in the first table) and simply change the spacings between numbers on the $y$y axis. That is, make the spacings between numbers proportional to the logarithms of the $y$y values.
Suppose for example we rule up the $y$y axis in the following way:
The first interval ( say of arbitrary length of $4$4 cm) starts from the origin, and covers the $y$y values from $1$1 to $10$10 ($10^0-10^1$100−101). The next $4$4 cm covers $y$y values from $10$10 to $100$100 ($10^1-10^2$101−102). The next $4$4 cm covers $y$y values from $100$100 to $1000$1000 ($10^2-10^3$102−103). The pattern continues with each $4$4 cm interval covering the $y$y values from $10^k$10k to $10^{k+1}$10k+1.
It is important to understand that within any of these intervals, the scale is not linear. Here is the beginning of the scale showing the first two intervals and the position of the first data point.
Note carefully that the gaps are getting smaller and smaller between $1$1 and $10$10 and between $10$10 and $100$100. Each tick between $1$1 and $10$10 is the position of $1,2,3,...9,10$1,2,3,...9,10. Each tick between $10$10 and $100$100 is the position of $10,20,30,...90,100$10,20,30,...90,100.
Because $\log_{10}20=1.301$log1020=1.301, the height of the point shown using the cm ruler would be $1.301\times4=5.204$1.301×4=5.204 cm.
Technically speaking, the above scale is called a semi-log $y$y scale because the $x$x-axis is still a linear scale. If we had changed the $x$x axis to a log scale instead of the $y$y -axis (for example if the $x$x values rather than the $y$y values had a large range), we would call it a semi-log $x$x scale. If we put both axes to log scales, we would call it a log-log scale.
On semi-log $y$y paper, a graph of the function $y=a^x$y=ax becomes a straight line. For example, consider the curve given by $y=2^x$y=2x for $x\ge0$x≥0. First we'll create a table of values shown here:
x | 0 | 1 | 2 | 3 | 4 | 5 |
---|---|---|---|---|---|---|
y | 1 | 2 | 4 | 8 | 16 | 32 |
We can plot the points on a semi-log $y$y graph as follows:
This is because, by taking logs on both sides, we see that $\log_{10}y=\log_{10}\left(2^x\right)=x\log_{10}\left(2\right)$log10y=log10(2x)=xlog10(2).
If we then set $Y=\log_{10}y$Y=log10y and the constant $\log_{10}2=m$log102=m, then the last equation becomes the straight line given by $Y=mx$Y=mx.
Generalising a little, the function $y=A\times2^x$y=A×2x will also be a straight line on semi-log $y$y paper since by taking logs, we have $\log_{10}y=\log_{10}A+x\log_{10}2$log10y=log10A+xlog102, which could be expressed as $Y=mx+c$Y=mx+c.
The important point being is that using semi-log paper must necessarily change familiar curve shapes to quite different shapes.
As a final note, plotting with semi-log scales is a common strategy used by scientists to verify to nature of certain collected data.
For example, it may be that a scientist looks at population data that seems to exhibit exponential growth. When the data is plotted on normal axes, it looks to rise in a way consistent with such a model.
To test the hypothesis, she might plot the data on semi-log $y$y paper to see if all the data points fall onto a straight line. If the data does, then she has verified that growth is indeed exponential.
Below is a table of values that shows a log scale relating $x$x and $y$y. Form an equation relating $x$x and $y$y. Express the equation in logarithmic form.
log scale measure ($y$y) | linear measure ($x$x) | |
---|---|---|
$0$0 | $=$= | $1$1 |
$1$1 | $=$= | $10$10 |
$2$2 | $=$= | $100$100 |
$3$3 | $=$= | $1000$1000 |
$4$4 | $=$= | $10000$10000 |
The decibel scale, used to record the loudness of sound, is a logarithmic scale. The lowest audible sound, with intensity $10^{-12}$10−12 watts/m^{2} is assigned the value of $0$0. A sound that is $10$10 times louder than this is assigned a decibel value of $10$10. A sound $100$100 ($10^2$102) times louder is assigned a decibel value of $20$20, and so on. In general, an increase of $10$10 decibels corresponds to an increase in magnitude of $10$10.
If the sound of a normal speaking voice is $50$50 decibels, and the sound in a bus terminal is $80$80 decibels, then how many times louder is the bus terminal compared to the speaking voice?
Give your final answer as a basic numeral, not in exponential form.
The Richter Scale is a base-$10$10 logarithmic scale used to measure the magnitude of an earthquake, given by $R=\log_{10}x$R=log10x, where $x$x is the relative strength of the quake. This means an earthquake that measures $4.0$4.0 on the Richter Scale will be $10$10 times stronger than one that measures $3.0$3.0.
The aftershock of an earthquake measured $6.7$6.7 on the Richter Scale, and the main quake was $4$4 times stronger. Solve for $r$r, the magnitude of the main quake on the Richter Scale, to one decimal place.