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1.1StatisticsStretch

Exponential Models

Many real datasets follow an exponential rather than linear pattern. Taking logarithms turns these models into straight lines, so linear regression can be applied to the transformed data.

26 min Regression, Correlation and Hypothesis Testing
Edexcel A Level Maths: 1.1 Exponential ModelCurated by Kepler, not produced by us. Watch it before the notes below.
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What you'll be able to do

  • Recognise exponential models y = abˣ and y = axⁿ
  • Linearise using logarithms
  • Interpret the gradient and intercept
  • Recover the original constants
1

Linearising y = abˣ

Taking logs of y=abxy = ab^x gives log⁡y=log⁡a+xlog⁡b\log y = \log a + x\log b — linear in xx with gradient log⁡b\log b and intercept log⁡a\log a.

y=abx ⇒ log⁡y=log⁡a+(log⁡b)xy = ab^x \ \Rightarrow\ \log y = \log a + (\log b)x
Plot log y against x.
2

Linearising y = axⁿ

For y=axny = ax^n, taking logs gives log⁡y=log⁡a+nlog⁡x\log y = \log a + n\log x — linear in log⁡x\log x with gradient nn and intercept log⁡a\log a.

Example 1A model y=axny = ax^n gives log⁡y=0.5+2log⁡x\log y = 0.5 + 2\log x. Find n and a.
1Gradient =n=2= n = 2.
2Intercept log⁡a=0.5⇒a=100.5≈3.16\log a = 0.5 \Rightarrow a = 10^{0.5} \approx 3.16.
Answer
n=2n = 2, a≈3.16a \approx 3.16.

Tip — y = abˣ: plot log y vs x. y = axⁿ: plot log y vs log x.

Formula recap

log⁡y=log⁡a+(log⁡b)x\log y = \log a + (\log b)x
For y = abˣ.
log⁡y=log⁡a+nlog⁡x\log y = \log a + n\log x
For y = axⁿ.

Common mistakes to avoid

Plotting log y vs log x for y = abˣ.
For abˣ plot log y vs x (only y is logged).
Reading a directly as the intercept.
The intercept is log a — undo with a = 10^{intercept}.

Key takeaways

  • y = abˣ → log y = log a + (log b)x (log y vs x).
  • y = axⁿ → log y = log a + n log x (log y vs log x).
  • Recover a from a = 10^{intercept}.

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

  1. For y=abxy = ab^x, plot log y against:

    • xx
    • log⁡x\log x
    • yy
    • x2x^2
    Show answer and explanation

    Answer: xx

    Only y is logged.

  2. For y=axny = ax^n, the gradient of log y vs log x is:

    • nn
    • log⁡a\log a
    • aa
    • log⁡n\log n
    Show answer and explanation

    Answer: nn

    Power becomes gradient.

  3. The intercept of the linearised model gives:

    • log⁡a\log a
    • aa
    • bb
    • nn
    Show answer and explanation

    Answer: log⁡a\log a

    Undo with a = 10^{intercept}.

  4. If log⁡y=0.5+2log⁡x\log y = 0.5 + 2\log x, then aa is about:

    • 3.163.16
    • 0.50.5
    • 22
    • 1010
    Show answer and explanation

    Answer: 3.163.16

    10^{0.5}.

  5. Why take logarithms of an exponential model?

    • to make it linear
    • to make it bigger
    • to remove x
    • to add noise
    Show answer and explanation

    Answer: to make it linear

    Then use linear regression.

Next step

Check your understanding

Work through questions on Exponential Models. Use the feedback to decide what needs another look.

3 past-paper questions · 6 marks

Practise this topic

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