12.5 - Logarithmic graphs

In section 6.1 we learned about the cartesian plane, rectangular coordinates, graphs and how to identify points on graphs. Before reading this section you may want to review that section.

Henceforth we will call the graphs described in section 6.1 linear graphs. In this section we will learn how to plot quantities on a new type of graph called a logarithmic graph. We will see how this can make it easier to determine the functional relationship between certain quantities.

We will motivate logarithmic graphs by giving two examples.

Example 1: Let t represent time and let i represent the electric current flowing in some electric circuit. The data shown in the table was collected for t and i. (For later use, the natural logarithm of the entries in column 2 was taken and entered in column 3.)

Find an equation relating i and t.

Solution: First we try plotting i versus t. But this only produces a curve as shown to the right and from this curve it is difficult to find the equation.

Next we try plotting ln (i) versus t. This does produce a straight line as shown to the right and we know how to find the equation of a straight line.

To find the equation of this line we will pretend for a moment that the axes are labelled y and x. Then this line must have the equation y = m x + b, where m and b are yet to be determined. Eventually we’ll put back the proper label ln (i) for y and t for x.

To determine m and b we will use the elimination method discussed in section 6.2. Let’s use these two points which lie on the line:

(x = 2.00 , y = 0.94) and (x = 8.00, y = −1.05).

Substituting the first point into the equation of the straight line, y = m x + b, gives the equation

0.94 = 2 m + b.

Substituting the second point into the equation of the straight line, y = m x + b, gives the equation

−1.05 = 8 m + b.

This pair of equations,

0.94 = 2 m + b
−1.05 = 8 m + b

is a system of two equations for the two unknowns m and b. We can solve them by elimination. Subtracting the first equation from the second equation eliminates b and gives:

−1.99 = 6 m m = −0.33

Back-substituting this value of m into, say the first equation, 0.94 = 2 m + b, then gives 0.94 = (2)(−0.33) + b which gives b = 1.6. So the equation of the straight line is

y = −0.33 x + 1.6.

Now replace back y by ln(i) and x by t:

ln (i) = −0.33 t + 1.6.

Antilogging this equation gives the equation

i = e^{−0.33 t + 1.6},

or simplifying,

i = 5 e^−0.33 t.

The key lesson learned from this example is that the graph of any exponential function y = a e^b x is a straight line when we plot ln (y) versus x.

Example 2: The curve to the right could be the graph of y = x², y = x³ or y = x⁴. From a plot of y versus x it is difficult to tell which it is.

But if we plot ln (y) versus ln (x) then these three functions become straight lines. These functions are called power functions because the variable x is raised to some power.

The key lesson learned from this example is that the graph of any power function y = a x^b is a straight line when we plot ln (y) versus ln (x).

Semilog And log-log graph paper

The figure above shows a logarithmic scale of the kind used on logarithmic graph paper. It is not a linear scale because larger numbers are allotted less space than smaller numbers. The lower part of the picture shows the logarithmic scale in more detail. The numbers along the axis are located where their logarithms would be placed on linear graph paper. This means that we can plot x itself on logarithmic graph paper rather than plot log (x) on linear graph paper. We avoid the job of taking logarithms on the calculator. Notice that moving to the left along a logarithmic axis only makes the numbers smaller and smaller. Zero is located an infinite distance to the left and negative numbers do not exist. The range from one power of 10 to the next is called a cycle.

Two types of logarithmic graphs are useful:

the semi-log graph, which has a logarithmic vertical scale and a linear horizontal scale, as shown below. the log-log graph, which has a logarithmic vertical scale and a logarithmic horizontal scale, as shown below.

The equations of straight lines on logarithmic graph paper

One purpose of logarithmic graph paper is simply to put wide ranges of data on one graph. Another purpose is to quickly check if a function follows an exponential law or a power law. We saw one example showing that plotting ln (y) versus ln (x) caused power law functions of the form y = a x^b to become straight lines. And we saw another example showing that plotting ln (y) versus x caused an exponential function of the form y = a e^b x to become a straight line. From these two examples we conclude the following:

A straight line on a semilog graph of y versus x represents an exponential function of the form
y = a e^b x.
A straight line on a log-log graph of y versus x represents a power law function of the form
y = a x^b.

To find the constants a and b, we can substitute two widely-spaced points which lie on the line into the appropriate equation. This gives two equations for the two unknowns a and b which can be solved by elimination.

Example: Find the equation of the straight line in the graph to the right.

Solution: A straight line on a log-log graph of Q versus T represents the power law function

Q = a T ^b.

We must find the values of a and b. To do this we will use a variation of the method described in section 6.2.

Take the two points shown in the graph and substitute them, one at a time, into the equation Q = a T ^b. This gives two equations in the two unknown constants a and b:

Here is the variation: first take natural logarithms of both equations

and then subtract the two equations. This eliminates ln (a). Then solve for b:

Note that b is the power of the power function and is often called the “slope of the log-log graph” and this equation is often used as a shortcut to compute it.

Back-substituting b into either of the previous equations gives ln (a) = 0.7120, and anti-logging gives a = 2.04. The constant a is often called the “intercept” although it occurs at T = 1, not at T = 0. (Remember that 0 does not exist in a logarithmic scale.)

So to 3 sig. figs. the equation describing the line is

Q = 2.04 t^0.558.

Example: Find the equation of the straight line in the graph to the right.

Solution: A straight line on a semi-log graph of P versus t represents the exponential function

P = a e^b t.

We must find the values of a and b. To do this we will use a variation of the method described in section 6.2.

Take the two points shown in the graph and substitute them, one at a time, into the equation P = a e^b t. This gives two equations in the two unknown constants a and b:

Here is the variation: first take natural logarithms of both equations

and then subtract the two equations. This eliminates ln (a). Then solve for b:

Note that b is the rate of the exponential function and is often called the “slope of the semilog graph” and this equation is often used as a shortcut to compute it.

Back-substituting b into either of the previous equations gives ln (a) = 4.377, and anti-logging gives a = 79.6. Note that the constant a is the “y intercept”. So to 3 sig. figs. the equation describing the line is

P = 79.6 e^−0.461 t.

Algebra Coach Exercises

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the semi-log graph, which has a logarithmic vertical scale and a linear horizontal scale, as shown below.	the log-log graph, which has a logarithmic vertical scale and a logarithmic horizontal scale, as shown below.