Log-log graphs and power laws

Why it works

Start from y = kxⁿ and take logs of both sides:

log y = log k + n log x

Rewrite it in the order of y = mx + c:

log y = n log x + log k

so plotting log y (y-axis) against log x (x-axis) gives a straight line with gradient n and y-intercept log k.

Explaining it in an exam

Examiners want:

• the log expansion written correctly;
• an explicit comparison with y = mx + c, with the terms in the same order and the + and = signs written;
• the words 'the gradient is n' (not just 'm') and 'the intercept is log k'.

Arrows or loops linking terms are not credited.

Analysing the graph

• Process both variables with the same base, to 3 d.p.
• Axis labels: log (y / unit) and log (x / unit).
• Gradient from a large triangle on the line; it has no unit.
• The intercept is read at log x = 0. If the axis does not include log x = 0, calculate it: c = log y - n log x for a point on the line.
• k = 10^c (the antilog); do not report the intercept itself as k.

Examples: L = L_sun (M / M_sun)^r for main-sequence stars; T = 2π sqrt(l/g) gives n = 1/2; an inverse-square law gives n = -2.

Worked example

Two points on a line of log y against log x are (0.200, 0.950) and (1.100, 2.300). Find n and k in y = kxⁿ.

1. n = gradient = (2.3 - 0.95) / (1.1 - 0.2) = 1.50
2. c = 0.95 - 1.50 x 0.2 = 0.650
3. k = 10⁰.650 = 4.467

Answer: n = 1.5, k = 4.47

Check yourself

1. Points on a log-log line are (0.300, 1.240) and (0.900, 2.440). Find the power n.
   Show answer

   2.0. n = 1.200 / 0.600 = 2.0.

2. That line has gradient 2.0 and passes through (0.300, 1.240). Find the intercept log k, to 3 d.p.
   Show answer

   0.640. c = 1.240 - 2.0 x 0.300 = 0.640.

3. The intercept is log k = 0.640. Find k, to 3 s.f.
   Show answer

   4.37. k = 10⁰.640 = 4.37.

4. A graph of log (T / s) against log (l / m) passes through (-0.500, 0.052) and (0.000, 0.302). Find the gradient.
   Show answer

   0.50. Gradient = 0.250 / 0.500 = 0.50, as expected for T proportional to sqrt(l).

5. The intercept is log k = 0.302, where T = k l⁰.5 and k = 2π / sqrt(g). Find g in m/s², to 3 s.f.
   Show answer

   9.83 m/s². k = 10⁰.302 = 2.0045; g = (2π / k)² = 9.83 m/s².

6. For main-sequence stars, a graph of log (L / W) against log (M / M_sun) has intercept 26.58 and gradient 3.5. Find L_sun, in units of 10²⁶ W, to 3 s.f.
   Show answer

   3.80 x 10²⁶ W. L_sun = 10²⁶.58 = 3.80 x 10²⁶ W.

7. Using L = L_sun (M / M_sun)³.5 with L_sun = 3.80 x 10²⁶ W, find L for a star of 2.0 solar masses, in units of 10²⁷ W, to 3 s.f.
   Show answer

   4.30 x 10²⁷ W. L = 3.80 x 10²⁶ x 2.0³.5 = 4.30 x 10²⁷ W.

8. What is the unit of the gradient of a log-log graph?
   • The unit of y divided by the unit of x
   • The unit of y
   • It has no unit
   • m
   Show answer

   It has no unit. Logs are pure numbers, so the gradient is too.

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Sources

• Pearson Edexcel International Advanced Subsidiary/Advanced Level in Physics specification (XPH11/YPH11), Issue 3
• Pearson qualifications: International Advanced Levels Physics (2018)
• Pearson examiners' report: IAL Physics WPH16/01, June 2024
• Pearson examiners' report: IAL Physics WPH16/01, October 2024
• Pearson examiners' report: IAL Physics WPH16/01, June 2023

Last updated 8 November 2026. Spotted a mistake? Tell us.