CAIE A-Level · Mathematics 9709 · Logarithmic and Exponential Functions

Reducing a Relationship to Linear Form (9709 Pure Mathematics 3)

9 min readSyllabus 3.2PreviewBy Uzair Khan

Syllabus objective

Use logarithms to transform a given relationship to linear form, and hence determine unknown constants by considering the gradient and/or intercept. E.g. y = kxⁿ gives ln y = ln k + n ln x, which is linear in ln x and ln y; y = k(aˣ) gives ln y = ln k + x ln a, which is linear in x and ln y.

Introduction

Many real-world relationships between variables are not linear — they follow power laws or exponential growth/decay. In 9709 examinations, you are frequently given a table of experimental data and asked to find the constants in an assumed model such as y=kxny = kx^n or y=kaxy = k a^x. Plotting yy against xx directly gives a curve, which is difficult to analyse precisely. However, by taking logarithms and applying the laws of logarithms, both types of relationship can be transformed into an equation of the form Y=mX+cY = mX + c — a straight line. Once you have a straight line, the gradient and yy-intercept uniquely determine the unknown constants. This technique appears regularly in Paper 3 and rewards careful algebraic manipulation.

Core Concept

The central idea is to linearise the relationship by taking natural logarithms (ln) of both sides, then use the laws of logarithms to separate the variables and constants into the form of a straight-line equation Y=mX+cY = mX + c.

Type 1: Power law — y=kxny = kx^n

Taking ln⁡\ln of both sides:

ln⁡y=ln⁡(kxn)=ln⁡k+ln⁡xn=ln⁡k+nln⁡x\ln y = \ln(kx^n) = \ln k + \ln x^n = \ln k + n\ln x

This can be written as:

ln⁡y⏟Y=n⏟mln⁡x⏟X+ln⁡k⏟c\underbrace{\ln y}_{Y} = \underbrace{n}_{m} \underbrace{\ln x}_{X} + \underbrace{\ln k}_{c}

So plotting ln⁡y\ln y (vertical axis) against ln⁡x\ln x (horizontal axis) gives a straight line with:

  • Gradient =n= n
  • Vertical intercept =ln⁡k= \ln k, so k=einterceptk = e^{\text{intercept}}

Type 2: Exponential law — y=kaxy = k a^x

Taking ln⁡\ln of both sides:

ln⁡y=ln⁡(kax)=ln⁡k+xln⁡a\ln y = \ln(k a^x) = \ln k + x\ln a

This can be written as:

ln⁡y⏟Y=ln⁡a⏟mx⏟X+ln⁡k⏟c\underbrace{\ln y}_{Y} = \underbrace{\ln a}_{m} \underbrace{x}_{X} + \underbrace{\ln k}_{c}

So plotting ln⁡y\ln y (vertical axis) against xx (horizontal axis) gives a straight line with:

  • Gradient =ln⁡a= \ln a, so a=egradienta = e^{\text{gradient}}
  • Vertical intercept =ln⁡k= \ln k, so k=einterceptk = e^{\text{intercept}}

Key point: You always take ln⁡\ln of both sides, never log⁡10\log_{10}. Either base works in principle, but natural logarithms are standard in 9709 and keep the algebra clean.

Unlock the full Logarithmic and Exponential Functions note with Nova

You're reading the preview. Unlock the complete note — every worked example, examiner pitfall and practice question — plus 24/7 AI tutoring from Nova that teaches directly from these notes.

Unlock with Nova Explore CAIE A-Level tutoring →

Keep learning

Prerequisites: The Laws of Logarithms

Explore CAIE A-Level Mathematics tutoring →

View the full Mathematics syllabus →

Part of Novark's free CAIE A-Level Mathematics notes