AQA · A-Level · Mathematics

    Course version: 7357 · Version from Sept 2017

    F: Exponentials and logarithms: Use logarithmic graphs to estimate parameters in relationships of the form y = axⁿ and y = kbˣ, given data for x and y.

    When data follow y = axⁿ or y = kbˣ, plotting on logarithmic axes turns the curve into a straight line whose gradient and intercept give the unknown parameters.

      Study notes

      From the checked teaching notes for this specification point.

      The idea

      When data follow y = axⁿ or y = kbˣ, plotting on logarithmic axes turns the curve into a straight line whose gradient and intercept give the unknown parameters. For y = axⁿ, take logarithms: log y = log a + n log x, so plotting log y against log x gives gradient n and intercept log a, hence a = 10^intercept. For y = kbˣ, log y = log k + x log b, so plotting log y against x gives gradient log b and intercept log k, hence k = 10^intercept and b = 10^gradient. Use the same logarithm base throughout, usually base 10. Estimate the gradient from two well-separated points on the line of best fit, then convert back by raising the base to the gradient or intercept. Always interpret the parameters in context and comment on how well the line fits the plotted points.

      What earns the marks

      • Takes logarithms of the given relationship and rearranges into the form Y = mX + c, identifying which quantities to plot.
      • Plots the transformed data correctly, using log y against log x for y = axⁿ and log y against x for y = kbˣ.
      • Draws a line of best fit and estimates its gradient using two well-separated points on the line.
      • Interprets the gradient and intercept correctly: for y = axⁿ, gradient is n and intercept is log a; for y = kbˣ, gradient is log b and intercept is log k.
      • Converts the intercept back to the parameter by raising the logarithm base to that value, and states the final relationship with the estimated parameters.
      • Comments on the suitability of the model by referring to how closely the plotted points lie to the straight line.

      Common mistakes and how to put them right

      • Plotting y against x on logarithmic axes but reading the gradient as the parameter directly.

        Put it right: Convert the gradient back, for example b = 10^gradient for y = kbˣ.

      • Forgetting to convert the intercept from log a or log k back to a or k.

        Put it right: Raise the base to the intercept value.

      • Using points that are close together to find the gradient, which magnifies reading error.

        Put it right: Choose two points far apart on the line of best fit.

      Exam tips

      • Label the axes clearly as log y and log x, or log y and x, so the examiner can see which transformation you have used.
      • Show the substitution of your two chosen points into the gradient formula before evaluating, and keep the unrounded gradient for the conversion.
      • Finish by writing the relationship with numerical values for the parameters and, where the context is given, interpret what they mean.

      Try the lesson

      The start of the F: Exponentials and logarithms: Use logarithmic graphs to estimate parameters in relationships of the form y = axⁿ and y = kbˣ, given data for x and y. lesson, made from checked teaching notes for this specification point.

      Have a go first

      Data satisfy a relationship of the form y = axⁿ. Explain what is plotted on each axis so that the graph is a straight line, and state what the gradient represents.

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