GRAPHING WITH LOGARITHMIC PAPERGuelph, Ontario, Canada http://www.physics.uoguelph.ca Go to this site to print free graph paper (en francais aussi): http://www.printfreegraphpaper.com/ 
At the end of the tutorial on Graphing Simple Functions, you saw how to produce a linear graph of the exponential function N = N_{0} e^{at} as shown in panel 1. This was done by taking the natural logarithm of both sides of the equation and plotting ln(N/N_{0}) vs t to get a straight line of slope a. 
Sometimes it's a nuisance to look up a bunch of logarithms of values of N/N_{0} so we make use of a special type of graph paper which does this automatically. A sheet of this paper is shown in panel 2. Notice that it has a linear scale horizontally but a logarithmic scale vertically. It's called "semilogarithmic paper". Notice that the vertical scale goes from 1 to 10. This paper is called "onecycle semilogarithmic paper". The significance of this name will become apparent in a little while. 
In panel 3, there's a table of values of N/N_{0} which obey an exponential relationship. In the righthand column, I've looked up the natural logarithms of N/N_{0} 
The graph shown in panel 4 is a plot of these values vs t. Notice that this graph is on normal graph paper, not semilog paper. We'll use semilog paper in a moment. As you see, the graph is a straight line and its slope, and thus the constant a, can be found. Pause for a moment and check the calculation of a. 
Now let's see how the semilog paper simplifies
all this
as shown in panel 5. The same data as in panel 3 is used here and,
since
our N/N_{0} data is all between 1 and 10, we can use the
numbers
on the lefthand edge of the graph paper just as they stand. All we
have
to do is plot the numbers as given. We don't have to find logarithms,
the
paper does it for us. That's the beauty of semilog paper. You have to
watch out how the paper is subdivided, though. In this example, it's
subdivided
in 0.1, from 1 to 3, but in divisions of 0.2, from 3 to 5 and 0.5 from
5 to 10. Pause here and see how this graph is plotted.
Now let's find a. Again, we must find the slope and this will involve finding logarithms but only at two points on whatever triangle we use to determine the slope. Pause again and check the calculation of the slope in panel 5. Notice, in fact, we had to look up only one logarithm in the
slope calculation
when we remembered that the difference of two logs is the log of the
quotient.
Of course, we got the same value as before,

Suppose, however, our data had been as shown in the table of panel 6. Now the values of t are the same but the values of N/N_{0} are 10 times larger. What do we do now? The answer is that the decade over which the vertical axis runs is quite arbitrary. It can be 1 to 10 as previously, or it can be 10 to 100 which is what we need now, or 100 to 1000, or 0.1 to 1, and so on. Pause and see that you understand how this graph in panel 6 was plotted. 
Now let's suppose that you have the data given
in the table
on panel 7. None of the semilog paper you have seen up to this point
will
work. You could plot the first number, or the 2nd to 5th, or the 5th to
7th, but you couldn't plot them all. Your onecycle paper will go only
from 1 to 10, or 10 to 100, or 100 to 1000, in other words, one decade.
But now N/N_{0} goes over parts of 3 decades, that is 1 to
1000.
For this you need threecycle semilog paper which has been
used
here to plot this data. Pause and check over the plot and calculation
on
panel 7.
As you can see from this, you choose the number of cycles in the graph paper you use to match the span of data which you have; semilog paper comes in one, two, three, seven cycles etc. 
Let's now turn to a new problem. Suppose you were presented with the set of data shown in panel 8. A graph of y vs x is also shown in panel 8, and you can see it's a smooth curve. But other than that, it's not very informative. Suppose, however, in addition, there were theoretical grounds for believing that this data obeyed a powerlaw, y = ax^{b}. How could we find if this were true and, if it were, evaluate the constants a and b? 
Let's take logarithms of both sides of the equation as in panel 9. You can see that, if y = ax^{b}, then a graph of log y vs log x yields a straight line of slope b and y intercept log a. Conversely, if a graph of log y vs log x for a set of data is a straight line, then the data does indeed follow the relation y = ax^{b}. Now we could look up a table of values of log x and log y and plot it but I won't bother to do it because, just as with the exponential law, there's a simpler way. 
Since we must plot log y vs log x, we need graph
paper divided
logarithmically along both axes. It's called
"loglog paper" and
a 1 x 1 cycle sample is shown in panel 10 where our data of panel 8 is
plotted. Pause and see that you understand how the points were plotted.
The graph is a straight line so the data does obey y = ax^{b}. Now let's find a and b. The constant b is given by the slope. Study panel 11 and see that you understand how the value was obtained. In calculating the slope, you may use either logs to the base 10 or logs to the base e as long as you are consistent. Notice that since logs have no units, then the slope has no units. The value of log a is the same as the value of the y intercept. To obtain this, we look on the graph for the point where the horizontal variable is 0. Remember that since the horizontal axis is logarithmic, the horizontal variable is actually log x, not just x, so we want the point where log x = 0. In order for log x to be 0, x must be 1. The y intercept can be read off the graph along the vertical line where x = 1. 
If values of x and y extend over more than one decade, then more cycles must be used. Loglog paper comes in many combinations, such as 2 x 1, 2 x 3 and 5 x 3.