1. Introduction

Many quantities in science can be described by equations of the form y=Axn. It is, though, not easy to distinguish between graphs of different power laws. Consider the data below:

It is not easy to see that the red points lie on a quadratic (y = Ax2) and that the blue data are on a quartic (y = Ax4). It is, however, clear that the black points lie on a straight line. Results from the modules on Logarithms and Straight Lines enable us to recast the power curves as straight lines and so extract both n and A.

Example 1: Consider the equation y = xn. This is a power curve, but if we take the logarithm of each side we obtain:

log (y) = log (xn)
= nlog (x) since log (xn) = nlog (x)

If Y = log (y) and X = log (x) then Y = nX. This shows the linear relationship. Plotting Y against X, i.e. log (y) against log (x), leads to a straight line as shown below.

Here n is the slope of the line. Thus from a log-log plot, we can directly read off the power, n.

Quiz 1: Which of the following lines is a log-log plot of y = x2?

(a) Line a Incorrect - please try again!
(b) Line b Correct - well done!
(c) Line c Incorrect - please try again!
(d) Line d Incorrect - please try again!

Explanation: The curve is y = x2. Taking logs of both sides gives:

log (y) = log (x2) = 2 log (x)

i.e. the log-log plot is a straight line through the origin with gradient 2.

Line b passes though the origin and through the point (x = 2, y = 4). From the modules on Straight Lines we know that the gradient, m, of a straight line passing through (log (x1), log (y1)) and (log(x2), log (y2)) is given by:

m = log (y2) − log (y1)/log (x2) − log (x2)

We see that the gradient of line b is given by:

m = 4 − 0/2 − 0 = 2

Therefore, line b is the correct log-log plot.

2. Straight Lines from Curves

Example 2: Consider the more general equation y = Axn. Again, we take the logarithm of each side:

log (y) = log (Axn)  
  = log (A) + log (xn)    since log (pq) = log (p) + log (q)
∴  log (y) = n log (x) + log (A)   since log (xn) = n log (x) 

The function log (y) is a linear function of log (x) and its graph is a straight line with gradient n which intercepts the log (y) axis at log (A).

Quiz 2: Referring to the lines a, b, c, d in the figure below, which of the following statements is not correct?

(a) If b corresponds to y = x3, then d would describe y = x−3. Incorrect - please try again!
(b) Lines a and c correspond to curves with the same power n. Incorrect - please try again!
(c) In the power law yielding c the coefficient A is negative. Correct - well done!
(d) If b is from y = x3, then in a the power n satisfies 0 < n < 3. Incorrect - please try again!

Explanation: If c corresponds to y = Axn, then log (y) = n log (x) + log (A). The intercept of line c on the log (y) axis is negative. This implies that log (A) < 0, which mean that 0 < A < 1. It does not signify that A itself is negative. (Of course, we also cannot take the logarithm of a negative number like this). It may be checked the other statements are correct.

Exercise 1: Produce log-log plots for each of the following power curves. In each case give the gradient and the intercept on the log (y) axis.

(a) y = x1/3

Solution: For y = x1/3, we get on taking logs: log (y) = 1/3log (x). This describes a line that passes through the origin and has slope 1/3. It is sketched below.

(b) y=10x5

Solution: For y = 10x5, we get on taking logarithms of each side: log (y) = 5 log (x) + log (10). This describes a line that passes through (0, log (10)) and has slope 5. It is sketched below.

(c) y=10x−2

Solution: The relation y = 10x−2 can be re-expressed as log (y) = −2 log (x) + log (10). This is sketched below.

(d) 1/3 x−3

Solution: If 1/3 x−3, then log (y) = −3 log (x) + log ( 1/3 ). This can also be written as log (y) = −3 log (x) − log (3). It is the equation of a line with slope −3 and intercept at −log (3). This line is sketched below.

Click on questions to reveal their solutions

Quiz 3: How does changing the base of the logarithm used (e.g., using ln(x) instead of log10 (x)), change a log-log plot?

(a) The log-log plot is unchanged. Incorrect - please try again!
(b) Only the gradient changes. Incorrect - please try again!
(c) Only the intercept changes. Correct - well done!
(d) Both the gradient and the intercept change. Incorrect - please try again!

Explanation: The equation of a log-log plot is:

log (y) = nlog (x) + log (A)

If we change the base of the logarithm that is used, then the gradient n is unchanged but the intercept A is altered.

For example, the log-log plot of y = 3x4 in terms of logarithms to the base 10 is:

log10 (y) = 4 log10 (x) + log10 (3)

- which has an intercept at log10 (3) = 0.477 (to 3 d.p.). Using natural logarithms the equation would become:

ln (y) = 4 ln(x) + ln (3)

This has the same gradient, but the intercept on the ln (y)-axis is now at ln (3) = 1.099 (to 3 d.p.)

The only exception to this is if A = 1, since logN = 0 for all N.

Note: In an equation of the form y = 5 + 3x2, taking logs directly does not help. This is because there is no rule to simplify log (5 + 3x2). Instead, we have to subtract the constant from each side. We then get y − 5 = 3x2, which leads to the straight line equation log (y − 5) = 2 log (x) + log (3).

3. Fitting Data

Suppose we want to see if some experimental data fits a power law of the form, y = Axn. We take logs of both sides and plot the points on a graph of log (y) against log (x). If they lie on a straight line (within experimental accuracy) then we conclude that y and x are related by a power law and the parameters A and n can be deduced from the graph. If the points do not lie on a straight line, then x and y are not related by an equation of this form.

Example 3: Consider the following data:

x 2 30 70 100 150
y 4.24 16.4 25.1 30.0 36.7

To see if it obeys y = Axn, we take logarithms of both sides. Here we use logarithms to the base 10. This gives the following table:

log10 (x) 0.30 1.48 1.85 2.00 2.18
log10 (y) 0.63 1.21 1.40 1.48 1.56

These points are plotted in the figure below.

It is evident that the red data points lie on a straight line. Therefore the original x and y values are related by a power law y = Axn. To find the values A and n, we first continue the line to the log10 (y)-axis which it intercepts at the blue dot: log10 (A) = 0.48. This means that A = 100.48 = 3.0 (to 1 d.p.).

The gradient of the line is esimated using two of the points

n = log (y2)−log (y1)/log (x2)−log (x1) = 1.56 − 0.63/2.18 − 0.3 = 0.5 (to 1 d.p.)

So the original data lies on the curve y = 3x1/2

Exercise 2:

(a) Rewrite y = 4x + 2 in such a way that it gives the equation of a straight line.

Solution: y = 4x + 4 can be re-expressed as follows:

Subtract 4 from each side:

y − 4 = 4x
y − 4 = 2 x
y − 4 = 2x1/2

Taking logarithms of each side yields

log (y − 4) = 1/2log (x) + log (2)

Thus plotting log (y−4) against log (x) would give a straight line with slope 1/2 and intercept log (2) on the log (y−4)-axis.

(b) What is the difference between two power laws if, when they are plotted on a log-log graph, the gradients are the same by the log (y) intercepts differ by log (3)?

Solution: If y = Axn then the log-log plot is the graph of the straight line

log (y) = nlog (x) + log (A)

So if the slope is the same, the power n is the same in each case.

If the coefficients A1 and A2 differ by

log (A1) − log (A2) = log (3)
log ( A1/A2 ) = log (3), since log (p/q) = log (p) − log (q)

So it follows that the coefficients are related by A1 = 3 A2

(c) Produce a log-log plot for the following data, show it obeys a power law and extract the law from the data.
 x  5  15  30  50  95
 y  10  90  360  1000  3610

Solution: To see if it obeys y = Axn, we take logarithms to the base 10 of both sides. The table and graph are below.

 log10 (x) 0.70 1.18 1.48 1.70 1.98
 log10 (y) 1 1.95 2.56 3.00 3.56

The data points are fitted by a line that intercepts the log10 (y) axis at log (A) = −0.40, so A = 10−0.40 = 0.4. The gradient can be calculated from n = (3 − 1) / (1.70 − 0.70) = 2. So the data lie on the straight line y = 0.4x2.

Click on questions to reveal their solutions

4. Quiz on Log-Log Plots

Choose the solutions from the options given

1.What are the intercept and slope respectively of the log-log plot of y=0.5x2?

(a) 1/2 and log (2)
(b) −log (2) and 2
(c) log (2) and 2
(d) log (1/2) and log (2)

2.If the log (y)-axis intercept of the log-log plot of y = Axn is negative, which of the following is true?

(a) n < 0
(b) A = 1
(c) 0 < A < 1
(d) n = − A

3.The data below obeys a power law, y = Axn. Obtain the equation and select the correct statment.

x 5 15 30 50 95
y 4.47 7.75 10.95 14.14 19.49
(a) n = 3
(b) A = 3/2
(c) n = 4
(d) A = 1/2

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