All subjects · Mathematics · WMA13 (P3)

Using log graphs to find a model

Taking logs turns a power or exponential model into a straight line. The gradient and intercept then give the constants in the model.

y = axn

Take logs: log y = log a + n log x. This is Y = mX + c with Y = log y and X = log x.

Plot log y against log x: the gradient is n and the intercept is log a.

y = kbx

Take logs: log y = log k + x log b.

Plot log y against x: the gradient is log b and the intercept is log k.

Finding the constants

Read the gradient and intercept, then undo the logs: if the intercept on a log10 graph is 0.48, then a = 100.48. On a ln graph use e instead. Use the base the question uses.

Worked example

A graph of log10 y against log10 x is a straight line with gradient 1.5 and intercept 0.6. Given y = axn, find a and n.

  1. log y = log a + n log x, so n = 1.5.
  2. log a = 0.6, so a = 100.6 = 3.98.

Answer: a = 3.98, n = 1.5

Key idea: y = axn: log y against log x (gradient n). y = kbx: log y against x (gradient log b). The intercept gives log a or log k.

Check yourself

  1. Data are thought to fit y = axn. Which graph should give a straight line?
    • log y against x
    • y against log x
    • y against x2
    • log y against log x
    Show answer

    log y against log x. log y = log a + n log x is linear in log x.

  2. The graph of log10 y against log10 x is a straight line with intercept 0.48 and gradient 2.5. Given y = axn, find a to 3 significant figures.
    Show answer

    3.02. log10 a = 0.48, so a = 100.48 = 3.02.

  3. For a model y = axn, the graph of log10 y against log10 x is a straight line through (1, 2.7) and (3, 6.3). Find n.
    Show answer

    1.8. n is the gradient: (6.3 - 2.7)/(3 - 1) = 1.8.

  4. A straight line through (1, 2.7) and (3, 6.3) on a graph of log10 y against log10 x has gradient 1.8. Given y = axn, find a to 3 significant figures.
    Show answer

    7.94. The intercept is 2.7 - 1.8 × 1 = 0.9, so a = 100.9 = 7.94.

  5. y = kbt. A graph of log10 y against t is a straight line with gradient 0.12 and intercept 0.7. Find b to 3 significant figures.
    Show answer

    1.32. log10 b = 0.12, so b = 100.12 = 1.32.

  6. For the model y = kbx, what does the gradient of the graph of log y against x represent?
    • log b
    • b
    • log k
    • k
    Show answer

    log b. log y = log k + x log b, so the gradient is log b.

  7. A population is modelled by P = 200(1.05)t. Find the gradient of the graph of log10 P against t, to 3 significant figures.
    Show answer

    0.0212. log10 P = log10 200 + t log10 1.05, so the gradient is log10 1.05 = 0.0212.

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Last updated 7 November 2026. Spotted a mistake? Tell us.