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 Video by Zeeshan Zamurred Regression, Correlation and Hypothesis Testing

Edexcel A Level Maths: 1.1 Exponential ModelWatch the full walkthrough 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

Linearising y = abˣ

Taking logs of $y = a b^{x}$ gives $lo g y = lo g a + x lo g b$ — linear in $x$ with gradient $lo g b$ and intercept $lo g a$ .

y = a b^{x} \Rightarrow lo g y = lo g a + (lo g b) x

Plot log y against x.

Linearising y = axⁿ

For $y = a x^{n}$ , taking logs gives $lo g y = lo g a + n lo g x$ — linear in $lo g x$ with gradient $n$ and intercept $lo g a$ .

1Gradient

= n = 2

2Intercept

lo g a = 0.5 \Rightarrow a = 1 0^{0.5} \approx 3.16

Answer

n = 2

a \approx 3.16

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

Formula recap

lo g y = lo g a + (lo g b) x

For y = abˣ.

lo g y = lo g a + n lo g 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

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