Logarithmic and Exponential Functions · 0606 Topic 6

Reducing Relationships to Linear Form

Teacher Rig, IGCSE Add Math tutor

Written by Teacher Rig

8 years teaching IGCSE Add Math · Updated 12 June 2026

Experimental data following y=axny = ax^n or y=Abxy = Ab^x plots as a curve, useless for reading off constants. Taking logs straightens it, and the constants reappear as gradient and intercept. This is one of 0606’s most predictable multi-mark questions, with a fixed three-step script.

The two standard reductions

Power model y=axny = ax^n: take logs of both sides:

logy=loga+nlogx\log y = \log a + n \log x Match to Y=mX+cY = mX + c: plot logy\log y against logx\log x \to gradient =n= n, intercept =loga= \log a

Exponential model y=Abxy = Ab^x:

logy=logA+xlogb\log y = \log A + x \log b Plot logy\log y against xx \to gradient =logb= \log b, intercept =logA= \log A

The diagnostic: a power model logs both variables; an exponential model logs only yy. (ln\ln works identically, with y=Aekxy = Ae^{kx}, plot lny\ln y against xx; gradient =k= k, intercept =lnA= \ln A.)

The three-step script

  1. Take logs and rearrange into straight-line form, show the log laws applied (M)
  2. State the matching: “plotting logy\log y against logx\log x gives a straight line with gradient nn and intercept loga\log a” (M/B, the statement itself scores)
  3. Extract the constants from the given/measured gradient and intercept

Step 3 hides the most-dropped mark in the subtopic: the intercept gives loga\log a, not aa. If the intercept is 0.60.6, then a=100.63.98a = 10^{0.6} \approx 3.98. Undo the log; don’t transcribe it.

Working from a given graph or table

Given a drawn line of logy\log y against logx\log x: gradient from two well-separated points =n= n; intercept (or substitution of one point) logaa\to \log a \to a. Given raw data: log the appropriate columns first, then fit. Either way, finish by writing the model with numbers in: y=3.98x1.5y = 3.98x^{1.5}, the assembled equation is usually the final A mark.

Common mistakes

  • Constants left as logs (a=0.6a = 0.6 instead of 100.610^{0.6})
  • The wrong variable logged for the model (both for power, only yy for exponential)
  • Matching statement skipped, constants pulled from nowhere score thinly
  • lg/ln mixed mid-question (pick one and stay)
  • Gradient read from adjacent, near-identical points (use the line’s full span)

This is the same linearising idea as the non-log conversions in straight-line graphs, one skill, two topics. Full topic context: Logs & Exponentials notes.

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