MATH 1441
Technical Mathematics for Biological Sciences
Semilogarithmic Graphs
This is the first of two documents in which we extend the basic ideas presented in the preceding document,
"The Equation of the Straight Line" to explore the construction, interpretation, and application of graphs in
which logarithmic scales are used. The material in this document is based very closely on the general
principles we've already described in connection with the graph and equation of a straight line. It would be
good if you understood those well (and so it might be worth reviewing that document before attempting to
work through this one). Even so, the discussion and examples given below should give you a deeper
understanding of those principles as well.
Just one caution here: the limitations of our graphics software (and our ability to use it) will mean that the
illustrations in this document are considerably less detailed, and certainly not identical in appearance to the
sort of graphs and graph paper you will use in class. We will attempt to make up some of this deficiency
with the discussion, but it will require some effort on your part to match the observations and calculations
given below with what you would actually encounter solving the same problems with pen and paper graphs.
Semi-logarithmic Graphs: The Strategy Behind the Graph
The problem addressed here is as follows: suppose you are presented with a set of observations in the
form of values of y for a sequence of values of t, and it is speculated that the relationship between y and t is
an exponential one:
y  a  er t
(1)
The goal here is in two steps:
(i)
to first confirm that it is reasonable to conclude that the relationship between the y-values and
the t-values really is formula (1), and
(ii)
if the answer to (i) is "yes", then, to obtain "best estimates" of the constants 'a' and 'r' in formula
(1) for this data.
Goal (i) will be accomplished by devising a way of graphing the data so that if formula (1) is the appropriate
relationship, the graph will turn out to be a straight line (or points apparently scattered uniformly along a
straight line path). We've already discussed the reasons why such an approach is necessary. Although
simply plotting the y values themselves might give us a curve that looks something like an exponential
curve, we can never positively distinguish one curved shape from other geometrically similar shapes (but
having very different mathematical formulas). The only "curve" which the human eye is really capable of
identifying unambiguously is the straight line, and so this is the shape of graph we need to rely on.
The second goal could be achieved rather trivially if we only had two observations. Then substituting those
two pairs of values of y and t into formula (1) would give us a system of two equations in the two unknowns,
'a' and 'r', which could be easily solved using the method already described in the course, and the problem
would be solved. However, in real life, few practitioners feel confident in experimental results based on just
two observations. However, once you have three or more observations, this simple approach doesn't work
in general. (The problem is that while in theory, the observations should all give points on the same straight
line, the presence of random observational errors that cannot be entirely eliminated in practice, means that
the observations will be scattered along the actual true straight line graph. This is true even when you just
have two observations, but you don't notice it because you can always find a straight line that passes
through two points, no matter how much in error they might really be!) It turns out that the straight-line graph
approach solves this problem automatically in a very effective way -- it really amounts to a method for
averaging out the random errors of observation.
The principle used to achieve the first goal is to first note that formula (1) can be rewritten in logarithmic
form:
David W. Sabo (1999)
Semilogarithmic Graphs
Page 1 of 11
ln y  r  t  ln a
(2)
Recognizing that 'r' and 'a' (and hence ln a) are constants for a particular exponentially growing or decaying
system, we see that formula (2) has the form of the equation of a straight line if we plot the value of t as the
horizontal coordinate, and the value of ln y as the vertical coordinate. Thus, for (y, t) data which satisfies
formula (1), and hence also the equivalent formula (2), a straight line graph will result if instead of plotting
the points at a vertical position proportional to the value of y, we plot the points at a vertical position
proportional to the value of ln y. The horizontal position remains proportional to the value of t.
Example 1: To test this idea, we can create some fake data which accurately satisfies a specific instance of
formula (1). Suppose r = 0.25/minute and a = 5, that is, that
y  6  e0.25t
Then, in a perfect world, if we were to take measurements of y when t = 0, 1, 2, 3, …, 10 minutes, we would
observe the following values:
Example 1
y
7.70
9.89
12.70
16.31
20.94
26.89
34.53
44.33
56.93
73.09
90
80
70
60
50
y
t
1
2
3
4
5
6
7
8
9
10
40
30
20
10
0
0
2
4
6
8
10
12
t (minutes)
The graph of this data is shown to the right above, and
you can see that it has the sort of shape that we expect of
a graph of exponential data plotted in the usual y vs. t format. A smooth exponential-looking curve passes
through all ten points (because we calculated the coordinates of those points exactly using the exponential
formula just above).
Now, we'll redraw this graph, but instead of plotting the points at a vertical position equal to their y-values,
we'll plot them at a vertical position equal to the
values of ln y. To do this, we must first extend our
Example 1
table of data to include the values of ln y.
6.0
y
7.70
9.89
12.70
16.31
20.94
26.89
34.53
44.33
56.93
73.09
ln y
2.042
2.292
2.542
2.792
3.042
3.292
3.542
3.792
4.042
4.292
5.0
4.0
ln y
t
1
2
3
4
5
6
7
8
9
10
3.0
2.0
1.0
0.0
0
2
4
6
8
10
12
t (minutes)
The resulting graph is shown above to the right, and
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Semilogarithmic Graphs
David W. Sabo (1999)
we see that the points do indeed fall exactly along a straight line. Both graphs above represent exactly the
same data. The shape is different because we have chosen to use a different approach to determine the
vertical positions of the points in the two graphs. What this simple example demonstrates so far is that if we
plot exponential data, (y, t), by placing the points at a vertical position proportional to the value of ln y and a
horizontal position proportional to the value of t, then the graph that results is a straight line. We call this
second graph a semilogarithmic graph, because one of the two axes (usually the vertical one) has a
logarithmically spaced scale, whereas the other axis (usually the horizontal one) still has the usual uniform
numerical scale.

Example 2: Consider, now, the following set of data, which is more typical of the actual sort of data one
might obtain for a potentially exponentially growing system:
t (minutes)
2.5
2.8
5.4
6.5
9.2
9.5
11.0
13.3
14.6
16.4
y
10.07
13.07
14.59
20.70
27.94
31.50
40.04
49.90
64.72
75.57
ln y
2.310
2.570
2.680
3.030
3.330
3.450
3.690
3.910
4.170
4.325
We've tabulated the values of ln y along with the (y, t) data values in the table. We get the following two
graphs from this data:
Exponential Growth: Example 2
Example 2
4.500
100.0
90.0
4.000
80.0
70.0
ln y
3.500
y
60.0
50.0
3.000
40.0
30.0
2.500
20.0
10.0
2.000
0.0
0.0
0.0
5.0
10.0
15.0
20.0
5.0
10.0
15.0
t
t
The graph on the left is suggestive of an exponential growth relationship between y and t, in that the points
seem to lie along an approximately exponentially-shaped curved path. However, it is quite clear from the
graph on the right that the points do lie scattered along a straight-line path. You can see from the rough line
sketched in that the pattern of points does not curve systematically in one direction or another.
Since the graph of ln y against t is a straight line, we know that the equation relating these two quantities
must be of the form of formula (2) above. But since formula (2) is equivalent to formula (1), we have
demonstrated that the data presented above is consistent with formula (1) -- the relationship between the y
values and the t values is exponential.

So, the strategy used to achieve the first goal in this problem is clear: Rather than plotting the y-values
vertically against the t-values horizontally, we plot the values of ln y vertically against the values of t
David W. Sabo (1999)
Semilogarithmic Graphs
Page 3 of 11
20.0
horizontally. If the plotted points appear to be scattered about a straight line path, then we are
justified in concluding that the data is consistent with an exponential relationship, formula (1).
If plotting ln y vs t gives a pattern of points which appears to curve in one direction or another, we know for
sure that the data is not consistent with formulas (1) and (2), and so the relationship between y and t is not
exponential.
The strategy used to achieve the second goal exploits properties of the equation of a straight line with
which you are very familiar. Having observed a straight line graph resulting from plotting the ln y values
vertically against the t values horizontally, we know that the equation of that straight line must be formula (2)
above. But this means that:

the coefficient, r, of t is the slope of the line. Thus, to calculate an estimate of r, we need
simply calculate the slope of the line.

when t = 0, we have y = a (or equivalently, ln y = ln a). Thus, if the line in the graph crosses
the position t = 0, we can read the value of ln a directly from the graph, from which we can
easily calculate the value of 'a'. If the line in the graph doesn't extend across t = 0 on the
graph we have, we can simply determine the (y, t) coordinates of any other point on the line
and substitute them into formula (1). Since we now have a value for r, the only unknown is 'a',
which can thus be easily computed.
It is important to keep in mind that these two sets of calculations are done using coordinates of points on the
straight line we have drawn, and not with the coordinates of any of the plotted points. You first plot the data
points individually. Then, using a ruler that is transparent enough for you to see all of the plotted points,
draw a line through the center of the pattern of points so that the pattern of points is separated into two
roughly equal parts, above and below the line. The idea here is that the line represents the pattern of points
you'd get if there were no random experimental errors occurred. We expect that the random errors would
show a relatively uniform pattern over the range of values of t for which observations were made. Thus, the
line should be drawn to avoid any systematic clustering of points above and/or below the line in particular
regions.
The line thus serves as an 'averaging" tool over all of the observations. All subsequent calculations are
based on the line itself and coordinates of points on the line, because the points on the line are considered
to be more accurate than any of the plotted points representing individual observations.
We will illustrate these calculations by continuing the two examples begun earlier.
Example 1 continued…
This is a good one to start out with because we
actually know what the values of r and 'a' are.
We need the coordinates of two points on the
line of the graph. For most accurate results,
these points should be as far apart as possible.
The best idea when you are doing these
graphs and calculations by hand is to select
the two points where the line leaves the grid
area of the graph paper. Not only are those
two points as far apart as possible, but it you
don't have to do any work to get two of the four
required coordinate values.
t = 12.0
ln y = 4.80
t = 0.0
ln y = 1.80
In this case, the line leaves the grid area of the
graph paper at t = 0 on the left and at t = 12 on
the right. When t = 0, we read ln y  1.80, and
when t = 12, we read ln y  4.80.
First, calculate the slope of the line:
Page 4 of 11
Semilogarithmic Graphs
David W. Sabo (1999)
r  slope 
actual rise
actual run
Now, since both scales of values in this graph are uniform scales, the actual distances in the vertical and
horizontal directions are just the differences between the coordinates of the two points. Thus
r
actual rise 4.80  1.80 3.00


 0.25
actual run
12.0  0.0 12.0
This is exactly the value of r that we previously used to set up the data to create the graph, so we have
consistency in this calculation.
In this case, we can obtain the coordinates of the point at which t = 0: they are (0, 1.80). Thus, we conclude
that at the level of accuracy with which we drew the graph,
ln a = 1.80
so
a  e1.80  6.05
This is in not too bad agreement with the known value a = 6 that we used to construct the original data. The
small discrepancy is due to the limited precision with which we were able to read coordinates from the
graph. With a really carefully drawn graph on millimeter ruled graph paper, we could have probably gotten
one additional decimal place of precision, which should have given very good agreement between the
values of a and r we started with, and the ones we computed from the graph constructed with the data.

t = 17.21
ln y = 4.500
Example 2, continued…
The graph of ln y vs t for Example 2 is shown
to the right again. The line that looks like the
best fit in our judgment leaves the grid area
at the points:
t = 0.00
ln y = 2.05
t = 0.0, ln y = 2.05
and
t = 17.21, ln y = 4.500
Thus, the computation of the slope, giving
the value of r is
r  slope 
actual rise 4.500  2.05

 0.1424
actual run
17.21  0
Again, we have a point on the line with t = 0, giving
ln a = ln y(t = 0) = 2.05
so that
a  e2.05  7.768
Thus, in summary, we can say that the data presented earlier in connection with this example is consistent
with an exponential growth model formula, and that our best estimate of that formula is
David W. Sabo (1999)
Semilogarithmic Graphs
Page 5 of 11
y  7.768  e0.1424t
Before leaving this example, we note that from the value of r, it is possible to compute an estimate of the
doubling time for this system:
t2 
ln2
ln2

 4.868
r
0.1424

Example 3: The following data was collected as part of a lengthy experiment to determine the persistence
of a genetic treatment in an experimental subject. Whenever blood samples were taken as part of ongoing
medical care and monitoring, the concentration of a particular molecule was determined. Sampling times, t,
are recorded in units of months, and the concentration, y, of the molecule of interest is in units of g per litre.
The twelve observations are as follows:
t (months)
1.3
3.6
5.3
8.5
11.7
13.4
t (months)
15.9
18.1
20.5
21.3
22.1
24.4
y (g/litre)
341.96
225.38
115.93
107.88
70.08
40.32
y (g/litre)
17.04
12.48
8.37
8.00
10.44
4.08
The problem here is to determine whether or not the concentration of this molecule decays exponentially
with time, and if so, what is the half-life of the concentration.
Solution
We know the strategy now. First, we plot the values of ln y vertically against values of t horizontally. If a
straight line plot results, we have confirmation that there is an exponential relationship, formula (1), between
y and t. The slope of that straight line will give us an estimate of the decay rate constant, r, from which we
can calculate the half-life of this process.
First, we must tabulate the natural logarithms of the y-values. This is easy to do, even if it is rather tedious:
y
341.96
225.38
115.93
107.88
70.08
40.32
17.04
12.48
8.37
8.00
10.44
4.08
ln y
5.835
5.418
4.753
4.681
4.250
3.697
2.836
2.524
2.125
2.079
2.346
1.406
Example 3
7.000
6.000
5.000
ln y
t
1.3
3.6
5.3
8.5
11.7
13.4
15.9
18.1
20.5
21.3
22.1
24.4
4.000
3.000
2.000
1.000
0.000
0.0
5.0
10.0
15.0
20.0
25.0
t
The graph of ln y vs. t is shown above
to the right. There is a fair degree of scatter in the pattern of the points, but it is fairly clearly scatter along a
straight line path. Thus, we conclude that this data is consistent with the relationship between y and t being
exponential. In fact, we can see from both the data and the graph that this must be a case of exponential
decay, because y is decreasing in value as time passes.
Page 6 of 11
Semilogarithmic Graphs
David W. Sabo (1999)
We've indicated with square boxes on the graph the two points where the straight line leaves the grid area of
the graph. These have the coordinates
t = 0.0
ln y = 6.07
t = 25.0
ln y = 1.42
and
(You won't be able to read the coordinates of these two points this precisely from the small illustration just
above in this document. However, if you constructed the graph on a standard sheet of millimeter-ruled
graph paper, you should be able to achieve this degree of precision.)
So now, we can calculate:
r  slope 
actual rise 6.07  1.42

  0.186
actual run
0.0  25.0
But, with this estimate of r, we can now compute
t1 
2
ln0.5
ln0.5

 3.727 months
r
0.186
The units of t1/2 arise from the fact that r has units of /month.
Thus, the concentration of the molecule of interest in this case has been confirmed to be decaying
exponentially with time, and we've been able to estimate the half-life of that concentration as approximately
3.727 months.
Incidentally, because we observed that ln y = 6.07 when t = 0, we can also compute
a  e6.07  433 g / litre
which completes the formula for y in terms of t, and also gives us an estimate of the concentration of this
molecule in the blood when the experiment began.

Using Semilogarithmic Graph Paper
What makes the methods described and illustrated above work is that the points are plotted so that their
vertical position is not proportional to the value of y, but rather, is proportional to the value of ln y. To
achieve, this, we had to calculate the natural logarithm of each observed value of y, and plot that value.
In the good old days before electronic calculators and computers, the determination of the natural logarithms
of the y-values was a fairly tedious step in an overall process that was even more tedious. People realized,
however, that this step of calculating logarithms could be avoided by using graph paper in which the
gridlines were drawn so that they were spaced in proportion to the values of the natural logarithms of
numbers. Particularly when semilogarithmic graphs are being prepared manually, the use of such paper is a
great convenience. Even though now people almost always use a computer application to prepare graphs,
the graphs are still often displayed with logarithmically-spaced gridlines, and so it is necessary for you to
know how to prepare such graphs, and how to analyze and interpret them to be able to get estimates of the
values of 'r' and 'a' in formula (1).
Just over the page is a simplified illustration of what semi-log-ruled graph paper might look like.
David W. Sabo (1999)
Semilogarithmic Graphs
Page 7 of 11
10
y
1
1
1
t
Commercially prepared graph paper of this type would have many more gridlines in both directions, but the
essentially features of the paper are shown by this diagram.
The horizontal scale is ruled with uniformly-spaced gridlines. The most commonly available graph papers of
this type have eight major gridlines one inch apart, and minor gridlines 0.1 inches apart in the horizontal
direction. We've shown just the major grid lines above.
You can easily see that the gridlines are not uniformly spaced in the vertical direction, but instead form
repeating bands of gridlines that start far apart and get closer together. Each such band (called a cycle)
corresponds to values between two consecutive powers of ten. Typically, the beginning of each cycle is
labeled with a 1, which you would replace by the appropriate power of 10 when you are planning your graph.
In the illustration above, we've just shown the major intermediate gridlines, which would typically be labeled
by consecutive integers: 2, 3, 4, 5, 6, 7, 8, and 9 (we omitted these labels in the figure above because of
lack of room). Usually, there are 5, 10, or even more minor gridlines between these major intermediate
gridlines depending on space available.
Now, when you plot points on this graph paper, you plot the actual y-values directly, using the numerical
scale values indicated with the non-uniform gridlines in the vertical direction. To start, you determine the
power of ten which brackets the y-values from below. For instance, if the smallest value of y in the data was
37, then the power of 10 bracketing the data from below would be 10 1 or 10. In most cases, this means that
you would relabel the lowest gridline with a '1' as '10'. Then, the next '1' up would be relabeled '100', the
next '1' up would be relabeled '1000' and so on. Further, the intermediate major gridlines are relabeled to
match. So the intermediate gridlines between 10 and 100 would be relabeled 20, 30, 40, 50, 60, 70, 80, and
90. The intermediate major gridlines between 100 and 1000 would be relabeled 200, 300, 400, 500, 600,
700, 800 and 900.
The point of all this relabelling is that now, when you plot the actual y-values according to this relabeled
scale, the vertical distances to the points will turn out to be proportional to the logarithms of the y-values.
Thus, without having to actually calculate the logarithms of the y-values, we will get a graph in which the
vertical distances to the points will be proportional to the logarithms of the y-values. By spacing the gridlines
in this non-uniform manner, the printer has really turn the graph paper into a logarithm calculator.
Just a few more remarks before we illustrate the use of such graph paper with the data in Examples 1 - 3
discussed previously. Unlike using uniformly ruled graph paper, we obviously cannot place our axes and
plot our graph just anywhere on logarithmically ruled paper. The gridlines labeled with 1's must always
correspond to successive powers of 10. Similarly, this means that you cannot "rescale" the vertical axis to fit
a broader range of values than the paper was designed to handle. The figure up above shows a paper with
three cycles, and that means that the y-values plotted have to fit within the range of three powers of ten.
Thus, this paper would be suitable if the y-values all fell into the range 1 to 1000, or 10 to 10000, or 0.01 to
10, etc. If your y-values ranged from near 0.01 to near 100, a span of four powers of ten, we would need
graph paper with four logarithmic cycles in the vertical direction. Logarithmically-ruled graph papers come
with a range of cycles, from as few as one cycle to perhaps as many as five cycles on standard sized sheets
of paper, and a person doing a lot of graphs on such paper manually would have to have a stock of paper
with a variety of cycles in order to accommodate varying ranges of values in the data.
Page 8 of 11
Semilogarithmic Graphs
David W. Sabo (1999)
Example 1 continued …
The data we used in example 1 is reproduced in the table below. The values of y range from a minimum of
7.70 to a maximum of 73.09. The first power of 10 less than 7.70 is 100 = 1, and the first power of 10
greater than 73.09 is 102 = 100. Thus, our vertical logarithmic scale must cover the interval 10 0 = 1 up to
102 = 100, two cycles in all.
t = 11.25
y = 100
t
1
2
3
4
5
6
7
8
9
10
y
7.70
9.89
12.70
16.31
20.94
26.89
34.53
44.33
56.93
73.09
80.0
70.0
60.0
50.0
40.0
30.0
20.0
10.0
It is most efficient to use two-cycle semilogarithmic
graph paper to plot this data. The lowest grid line
on the paper is labelled y = 1, the beginning of the
second cycle from the bottom is labeled y = 10,
and the top of the second cycle (which would be
the top gridline of two-cycle paper) would be
labeled y = 100. It's best to actually write these
labels directly on the paper to begin with. You
might also label the major intermediate gridlines in
the second cycle with the values 20, 30, 40, 50,
60, 70, 80 and 90. It should be quite clear how to
do this.
8.0
7.0
6.0
5.0
4.0
3.0
2.0
t = 0.0
y=6.0
1.0
You need to also set up a horizontal scale. Since the paper will likely have about seven major divisions
horizontally, and t varies from 1 to 10 here, it is probably most convenient to have each major division in the
horizontal direction correspond to an interval of 2 in the values of t, starting with t = 0 at the left edge of the
paper.
Now, simply plot the (t, y) values as points at the positions indicated by the two scales of numbers. By
plotting the values of y directly as indicated by the gridline labels, the vertical distances to each point will end
up being proportional to the logarithms of the y-values.
The sort of graph you end up with is shown in the figure above to the right. Because of limited space here,
we've omitted all of the minor gridlines, and some of the labels along the vertical axis. As in previous graphs
involving this data, the points all lie along the same straight line because the numbers themselves were
generated to fit a particular exponential growth formula exactly. This is our simple test case for the methods
explained in this document.
Getting a straight line graph in this way confirms that the relationship between y and t is an exponential
relationship, formula (1), because we have effectively positioned the points at vertical distances proportional
to ln y. We've reviewed the argument many times that a linear relationship between ln y and t is equivalent
to an exponential relationship between y and t.
What's left to do then is to compute the best estimates of the constants 'r' and 'a' in formula (1). The straight
line graph here confirms formula (2) is correct, indicating the slope of the line is the value of r. To calculate
the slope of the line, we read the coordinates of two widely separated points on the line. In the figure, we've
identified the two points where the line leaves the gridline area:
t = 0.0, y = 6.00
and
t = 11.25, y = 100.0
David W. Sabo (1999)
Semilogarithmic Graphs
Page 9 of 11
Thus,
r  slope 
actual rise ln100.0  ln6.00

 0.250
actual run
11.25  0.0
Notice that the actual rise here is computed as the difference of the natural logarithms of the two ycoordinates. That is the correct formula because the horizontal gridlines semilogarithmic paper are spaced
so that the actual vertical distances are proportional to the natural logarithms of the indicated scale values.
Thus, the distance between the horizontal gridlines labelled y = 6.00 and y = 100.0 is not the difference of
the y-values (as would be true with uniformly ruled graph paper), but is equal to the difference in the natural
logarithms of these two values.
From formula (1), a = y when t = 0. From the graph above, we see that y = 6.00 when t = 0, so we conclude
that a = 6.00 in this case. Thus, from the semilogarithmic graph plotted just above, we conclude that the
formula summarizing the data in the table above is
y  6.00  e0.250t

Example 2 continued …
The values of y in Example 2 above range
from a minimum of 10.07 to a maximum of
75.57, so that technically, one-cycle
semilogarithmic graph paper will suffice in
this case, with the cycle running from 10 to
100. The advantage of using a single-cycle
sheet of paper is greater accuracy (because
the region of the plotted points would occupy
a greater proportion of the sheet), but has the
disadvantage that the vertical intercept is now
off the graph, making the calculation of the
constant 'a' slightly more difficult.
In the graph illustrated to the right, we've
plotted the data in the upper cycle of a sheet
of two-cycle paper. The points appear to be
scattered uniformly about a straight-line path,
confirming that the relationship between y
and t in this data is exponential. We've noted
the coordinates of the two most widely
separated points on the graph:
t = 0.0  y = 7.74
t = 17.95
y = 100
100.0
90.0
80.0
70.0
60.0
50.0
40.0
30.0
20.0
t=0
y = 7.74
10.0
9.0
8.0
7.0
6.0
5.0
4.0
3.0
2.0
and
t = 17.95  y = 100.
From these values we can calculate the slope
of the line, giving the value of r:
r  slope 
1.0
0.0
5.0
10.0
15.0
20.0
actual rise ln100.0  ln7.74

 0.1425
actual run
17.95  0.0
Furthermore, one of the two points listed is the vertical intercept (since the t-coordinate is 0.0), and so we
can say here that a = 7.74. Thus, it appears from this data that the best approximation to a formula relating
y to t is
Page 10 of 11
Semilogarithmic Graphs
David W. Sabo (1999)
y  7.74  e0.1425t
If we had used single-cycle graph paper, we would have found that the point where the graph leaves the
region of grid lines at the bottom would have the approximate coordinates:
t = 1.80  y = 10.0
Now there is no visible vertical intercept from which we can read the value of 'a' directly. Instead we need to
compute a value for 'a' given our value for r, and selecting values for y and t from the graph. For example,
using t = 17.95, y = 100.0 and r = 0.1425, we get from formula (1) that
a
y
rt
e

100.0
 7.747
 0.1425 17.95 
e
which is quite close to the value we read off of the two-cycle graph directly.

Example 3 continued …
We need to say relatively little about solving
Example 3 using the semilogarithmicallyruled graph paper. The y-values in the data
range from a minimum of 4.08 to a
maximum of 341.96, so we really need
three-cycle semilogarithmic graph paper to
plot the data (since we need to cover the
range of y-values from y = 100 = 1 to
y = 103 =1000).
The resulting graph is illustrated to the right.
We've determined the coordinates of two
extreme points on the graph, from which
you can calculate that the slope of the line,
and hence the value of r is -0.1858. One of
those two points is the vertical intercept,
giving a = 431. This gives us the final
formula relating y to t as
t = 0.0
y = 431
100.0
10.0
y  431 e0.1858t
This is very similar to the result we
previously obtained using uniformly-ruled
graph paper.
t = 25.0
y = 4.14
1.0

David W. Sabo (1999)
Semilogarithmic Graphs
Page 11 of 11