MATH 1441
Technical Mathematics for Biological Sciences
Log-Log Graphs
We found that we could turn an exponential curve into a straight line graph by using a logarithmic scale
along one axis of the graph. It turns out that one other commonly encountered mathematical formula pattern
can be "linearized" by using logarithmic scales on a graph: the so-called power function
y  A  xn
(1)
where y and x are the variables, and A and n are constants. Examples of formulas of this form arise in a
variety of scientific and engineering applications.
Equating logarithms of both sides of formula (1) gives

log  y   log A  xn
or

 
log  y   log xn  logA
or
log  y   n  logx  logA
(2)
This equation has the form of the equation of a straight line if, instead of plotting y vertically vs. x
horizontally, we plot values of log y vertically vs. values of log x horizontally. The slope of that straight line
will be the value of the constant exponent n in formula (1), and from coordinates of points on that line, we
can also calculate the value of the constant A. In fact, A = y when x = 1 (or log x = 0).
We will illustrate the calculations involved with two examples.
Example 1: The table just below to the left gives some experimental data for which it is believed a formula
of the form (1) applies. A plot of y vs. x on ordinary
uniform axes scales is shown to the right of the table,
Example 1: y vs. x
and while it has a plausibly curved shape, provides no
certain proof that formula (1) is correct.
7000
y
160
300
350
710
900
1510
1900
3200
4000
5800
6000
5000
4000
y
x
0.15
0.25
0.72
1.50
3.55
6.1
15
30
60
90
3000
2000
1000
0
0
20
40
60
80
100
x
To verify that formula (1) really is the appropriate
formula for this data, we need to construct a plot of the
logarithms of the x and y values. Then, if a straight line results, we have confirmation that formula (2) is the
correct formula. However, since formula (2) is equivalent to formula (1), a straight line graph resulting from
the use of a logarithmic scale on both axes is confirmation that formula (1) is the correct formula.
As with semi-logarithmic graphs, we can proceed in one of two ways here. The simplest (but most tedious)
is to tabulate the values of the required logarithms, and plot the points on ordinary uniformly-ruled graph
David W. Sabo (1999)
Log-Log Graphs
Page 1 of 5
paper. We need to use a specific base b for our logarithms. Our choice is really a matter of convenience,
and for these examples, we'll choose to use base 10 or common logarithms.
So, we have
Example 1: Log10y vs Log10x
4.00
y
160
300
350
710
900
1510
1900
3200
4000
5800
log10x
-0.824
-0.602
-0.143
0.176
0.550
0.785
1.176
1.477
1.778
1.954
log10y
2.204
2.477
2.544
2.851
2.954
3.179
3.279
3.505
3.602
3.763
3.80
3.60
3.40
log10y
x
0.15
0.25
0.72
1.50
3.55
6.1
15
30
60
90
3.20
3.00
2.80
2.60
2.40
2.20
2.00
-1.00
You see from the graph sketched to the right that we
do get a pattern of points that appear to be scattered
along a straight line path. This confirms that the data
obeys a formula of the pattern in (1).
0.00
1.00
2.00
3.00
log 10x
To calculate the value of the exponent, n, from this graph, we need to compute the slope of the graph. this
requires reading the coordinates of two widely separated points on the line. From an accurately drawn
version of this graph on a sheet of paper, we found that the line left the gridded region of the paper at the
following two points:
log10x = -1.00

log10y = 2.170
log10x = 2.452

log10y = 4.00
and
Note that our notation here is correct  these numbers are values of log10x and log10y.
Since the grid is uniformly-spaced on this graph, we calculate the slope using the differences of these
coordinates to give distances vertically and horizontally:
n  slope 
rise
4.00  2.170

 0.530
run 2.452   1.00 
The line log10x = 0 on the graph corresponds to x = 1, and so the point at which the graph intersects this
vertical line has coordinates (0, log10A). Again, reading from a more accurately constructed graph, we get
that
log10A  2.700
so that
A  102.700  501
Thus, we conclude that the formula best describing the original data is:
y  501 x0.530
Of course, just as in the case of semi-logarithmic graphs, there is graph paper available commercially with
logarithmically-spaced gridlines. If we use that kind of paper, we could plot the x- and y-values directly
without having to first calculate their logarithms. The procedure for setting up such graphs is very similar to
that used in setting up semilogarithmic graphs on semilogarithmically-ruled graph paper, except that now we
need to devise appropriate scales for both the x-axis as well as the y-axis.
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Log-Log Graphs
David W. Sabo (1999)
For the present set of data, the x values range from a minimum value of 0.15 to a maximum value of 90.
Thus, our x-scale must range from 10-1 = 0.1 up to 102 = 100, a total of three cycles. The values of y range
from a minimum of 160 to a maximum of 5800, so the y-axis must span the values from 102 = 100 to
104 = 10000, a total of two cycles. Thus, we need graph paper with two logarithmic scales: the vertical scale
accommodating two cycles, and the horizontal scale
accommodating three cycles.
Example 1: y vs x (Logarithmic Scales)
Again (as expected), the plotted points appear to lie
along a straight line path, consistent with formula (2)
and hence formula (1) above. Thus, the exponent n in
formula (1) will be given by the slope of this straight
line. To calculate that slope, we first read the
coordinates of the two most widely separated points on
the graph. Using an accurately constructed graph on
paper for this problem, we obtained:
x = 0.100

y = 148
x = 100.0

y = 5760
10000
y
An illustration of the resulting graph is shown in the
figure to the right. We've left a fair amount of detail out
of this illustration, but you can see the typical
logarithmic variation in spacing between grid lines, now
along both the x-axis and the y-axis.
1000
100
0.10
1.00
x
10.00
100.00
and
Thus
n  slope 
log10 5760  log10 148
actual rise

 0.530
actual run log10 100.0  log10 0.100
which is exactly the same value we obtained before. Because of the way the paper is ruled in both the
horizontal and vertical directions, the actual horizontal and vertical distances between the two points are
obtained as the differences of the logarithms of the coordinates.
We can read the value of A right off the graph from the point where the graph crosses the vertical line
labeled x = 1. On our graph, this point appears to be right at the horizontal line labeled y = 500, so we
estimate that A = 500. This agrees well with the value 501 that we obtained earlier.
By the way, this final formula
y  501 x0.530
should seem quite plausible to you. The original graph of y vs x on uniformly-ruled graph paper produces a
pattern of points very similar to the graph of y  x  x0.5, which is approximately what our formula ended up
be. Obviously, it would be impossible to distinguish the difference between y  x0.50 and y  x0.53 from a
graph of just ten points with coordinates containing perceptible errors of measurement. Using a log-log
graph approach we can make such fine distinctions. (The issue of whether this distinction is real, rather
than just the result of random measurement errors is one we will have to defer consideration of until late
next term. One of the major goals of our study of statistics next term will be to come up with ways of
assessing how uncertain the results, such as the value of n here, obtained from experimental data may be.)

David W. Sabo (1999)
Log-Log Graphs
Page 3 of 5
Example 2:
We'll go through this second example rather quickly, since the basic ideas should all be fairly clear now.
The data consists of twelve observations of corresponding x and y values that are suspected of obeying a
power law formula:
y
9050
6950
3450
2470
1280
1240
790
608
306
248
163
149
log10x
-0.602
-0.398
-0.347
-0.161
-0.108
0.009
0.057
0.207
0.215
0.401
0.405
0.604
Example 2: y vs. x
log10y
3.957
3.842
3.538
3.393
3.107
3.093
2.898
2.784
2.486
2.394
2.212
2.173
10000
9000
8000
7000
6000
y
x
0.25
0.40
0.45
0.69
0.78
1.02
1.14
1.61
1.64
2.52
2.54
4.02
5000
4000
3000
2000
1000
0
0.00
1.00
2.00
x
3.00
4.00
5.00
The plot of y vs x on ordinary graph paper shown
above to the right is certainly not a straight line. Both from the data and this graph, y seems to be
decreasing with x, and so if formula (1) is the right type of formula for this data, we expect the exponent, n,
to have a negative value.
Using either the log values in the table above and uniformly-ruled graph paper, or the appropriate type of
log-log graph paper, we get the following two graphs:
Example 2: y vs. x (Logarithmic Scales)
10000
Example 2: Log10y vs Log10x
4.500
y
4.000
Log10y
3.500
1000
3.000
2.500
2.000
1.500
-0.800 -0.600 -0.400 -0.200 0.000 0.200 0.400 0.600 0.800
Log10x
100
0.10
1.00
x
Both plots show a pattern of points along a straight line path, and in both the slope of the line is negative
(indicating the value of n, as previously surmised, will be negative).
Starting with the graph on the left above, we find that the extreme points of the graph have coordinates
log10x = -0.800

log10y = 4.34
log10x = 0.800

log10y = 1.71
and
Thus
n  slope 
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rise
4.34  1.71

  1.644
run 0.800  0.800
Log-Log Graphs
David W. Sabo (1999)
10.00
and from the place where the graph crosses the vertical line log 10x = 0, we get
log10 A  3.03

A  103.03  1072
Thus, the formula fitting this data is apparently
y  1072  x1.644
To duplicate these calculations using the logarithmically-ruled graph, we use the coordinates of the extreme
points of the graph on the right above, which we find to be:
x = 0.256

y = 10000
x = 4.22

y = 10
and
Then
n  slope 
log10 10000  log10 10
actual rise

  1.643
actual run log10 0.256  log10 4.22
which is very similar to the result we obtained from the first graph. The value of A is the y-coordinate of the
point where the graph cuts the vertical line x = 1, for which we get A  1070.

One more brief comment is in order before leaving this topic and this document. You might wonder how
best to proceed if you have some experimental data, but not necessarily any good idea of the sort of formula
appropriate to that data. If you have no basic scientific or technical principles to guide you, then you more or
less just have to "guess" at the correct type of formula, and then check whether your guess is valid or not.
We know that the data for Example 2, above, for instance, gives a log-log plot which is pretty clearly a
straight line, verifying formula (2) and (1) as the appropriate one in this case.
Example 2: y vs. x (Semilogarithmic Test)
10000
y
If you had started out speculating that the data
in this example might obey an exponential
formula, then you would have prepared a
semilog graph, which would look something like
the graph to the right. In this case, the points
clearly are not scattered along a straight line
path (the line we drew in the figure has no
significance, other than to emphasize the
curvature of the pattern of the plotted points).
A plot like this is a clear indication that the
speculated exponential relationship is not
supported.
1000
100
0.00
David W. Sabo (1999)
Log-Log Graphs
1.00
2.00
3.00 x
4.00
5.00
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