Linearizing Data with Logarithms (AI HL)
A power or exponential relationship looks curved on a normal scatter plot, which makes it hard to fit by eye or estimate parameters from. Taking logarithms of the data straightens that curve into a line, so the ordinary tools of linear regression - gradient and intercept - can be used to recover the original model. It's part of the broader Exponential & Logarithmic Models topic.
31 questions on this sub-topic.
Two linearizing transforms
Covered under IB syllabus reference AHL2.10: linearizing data using logarithms to determine whether the data has an exponential or a power relationship, and using best-fit straight lines to determine the parameters. In examinations, students are not expected to sketch log-log or semi-log graphs by hand.
Power model \(y=ax^n\)
\(\log y = n\log x + \log a\)
Plot \(\log y\) against \(\log x\) (a log-log graph). The line's gradient gives \(n\); its \(y\)-intercept gives \(\log a\), so \(a=10^{\text{intercept}}\).
Exponential model \(y=ab^x\)
\(\log y = (\log b)x + \log a\)
Plot \(\log y\) against \(x\) itself (a semi-log graph). The gradient gives \(\log b\); the \(y\)-intercept gives \(\log a\).
Need the surrounding syllabus context and the parent's GDC walkthrough for finding lines of best fit? See Exponential & Logarithmic Models, including its GDC guidance.
Worked examples
Data follows the model \(y = ab^x\). Show that taking \(\log_{10}\) gives a linear relationship, and identify the gradient and \(y\)-intercept.
Worked solution
\(\log y = \log a + x \log b.\) M1
This is \(Y = (\log b)x + \log a\), linear in \(x.\) A1
Gradient \(= \log b\); \(y\)-intercept \(= \log a.\) A1
The table shows values of \(x\) and \(y\):
| \(x\) | 1 | 2 | 5 | 10 |
|---|---|---|---|---|
| \(y\) | 2 | 5.66 | 22.4 | 63.2 |
A power model \(y = ax^n\) is proposed. Linearise the data and use the first and last points to estimate \(n\) and \(a\).
(a)(i) State the value of \(n\).
(a)(ii) State the value of \(a\).
Worked solution
\(\log x\): 0, 0.301, 0.699, 1. M1
\(\log y\): 0.301, 0.753, 1.350, 1.801. A1
Gradient \(n = \dfrac{1.801 - 0.301}{1 - 0} = 1.5.\) M1
A1 Intercept \(\log a = 0.301 \Rightarrow a = 2.\) A1
Common mistakes
- Using the wrong pair of variables. A power model needs \(\log y\) plotted against \(\log x\); an exponential model needs \(\log y\) plotted against plain \(x\). Mixing these up gives a curve that still isn't straight.
- Reading the gradient as \(a\) and the intercept as \(n\) (or vice versa). For \(y=ax^n\), the gradient of the linearised line is the power \(n\); the constant \(a\) only appears after undoing the log on the intercept.
- Forgetting to undo the log on the intercept. The \(y\)-intercept of the linearised line is \(\log a\) (or \(\log b\)), not \(a\) itself - you must raise 10 (or \(e\), depending on which log was used) to that value to recover the original constant.
Ready to practise properly?
31 linearizing-data questions, marked instantly like the real exam.
Quick answers
Why do you take logarithms of data to fit a power or exponential model?
Taking logs turns a curved power or exponential relationship into a straight line, so an ordinary line of best fit can be used to estimate the model's parameters.
How do you recover the original parameters after linearizing a power model?
For \(y = ax^n\), plotting \(\log y\) against \(\log x\) gives a line \(\log y = n(\log x) + \log a\). The gradient is \(n\), and \(a\) is found by raising 10 (or \(e\)) to the power of the \(y\)-intercept.