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Notes/Mathematics/Paper 2/Logarithmic and Exponential Functions
CAIEAS Level9709§2.2

Logarithmic and Exponential Functions

Logarithms are the tool for getting at an unknown that is stuck in an exponent — and for straightening a curved law into a line you can measure.

34 min read 4 sub-topics
96
question parts
2021–2025 · 37 papers
8 marks
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≈ 16% of the paper
2.3/3
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#5
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Every operation in mathematics has an undo. Adding is undone by subtracting, squaring by square-rooting — and raising to a power is undone by taking a logarithm. That is the entire idea:

ax=b  ⟺  x=log⁡aba^x = b \iff x = \log_a b

Once you have that, the rest of the topic is bookkeeping: three laws for manipulating logarithms, one special base (e\mathrm{e}) that calculus prefers, and one genuinely clever application — taking logs of an experimental law to turn a curve into a straight line whose gradient and intercept you can actually read.

It carries about 8 of the 50 marks on every Paper 2, and the recorded techniques are blunt about where they come from: "take logarithms of both sides" and "combine logarithms into a single logarithm" between them account for most of the topic.

Before you start you should be able to

  • Laws of indices: aman=am+na^m a^n = a^{m+n}, aman=am−n\dfrac{a^m}{a^n} = a^{m-n}, (am)n=amn\left(a^m\right)^n = a^{mn}, a0=1a^0 = 1

  • Solve linear and quadratic equations

  • Inverse functions, and reflection in y=xy = x (see Functions)

  • The equation of a straight line, y=mx+cy = mx + c (see Coordinate Geometry)

By the end of this page you can

  • Convert between ax=ba^x = b and x=log⁡abx = \log_a b

  • Use the three laws of logarithms to combine or split expressions

  • Describe ex\mathrm{e}^x and ln⁡x\ln x as inverses, and sketch both, including y=ekxy = \mathrm{e}^{kx} for positive and negative kk

  • Solve equations and inequalities where the unknown is in the index

  • Reduce y=kxny = kx^n or y=kaxy = k a^x to linear form and find the constants from a gradient and an intercept

01

What a logarithm is, and its three laws

Syllabus requirement · §2.2

“

understand the relationship between logarithms and indices, and use the laws of logarithms (excluding change of base).

”

ax=b  ⟺  x=log⁡aba^x = b \iff x = \log_a b

A logarithm is an index

·

Read log_a b as “the power you raise a to, to get b”.

So log⁡28=3\log_2 8 = 3 because 23=82^3 = 8, and log⁡101000=3\log_{10} 1000 = 3 because 103=100010^3 = 1000. Every logarithm law is an index law wearing different clothes — which is why log⁡(mn)=log⁡m+log⁡n\log(mn) = \log m + \log n mirrors apaq=ap+qa^p a^q = a^{p+q}.

The three laws
log⁡a(mn)=log⁡am+log⁡an\log_a(mn) = \log_a m + \log_a n

Multiplication becomes addition

log⁡a ⁣(mn)=log⁡am−log⁡an\log_a\!\left(\frac{m}{n}\right) = \log_a m - \log_a n

Division becomes subtraction

log⁡a ⁣(mk)=klog⁡am\log_a\!\left(m^k\right) = k\log_a m

A power comes out to the front — the useful one

Values worth knowing on sight
log⁡a1=0\log_a 1 = 0

anything to the power 0 is 1

log⁡aa=1\log_a a = 1

a to the power 1 is a

ln⁡ex=xandeln⁡x=x\ln \mathrm{e}^x = x \quad\text{and}\quad \mathrm{e}^{\ln x} = x

each undoes the other

Two things are not in the syllabus

Change of base is excluded — you will never need log⁡ab=log⁡blog⁡a\log_a b = \dfrac{\log b}{\log a} in Paper 2. And log⁡(m+n)\log(m+n) does not simplify: there is no law for the log of a sum, and inventing one is the most common error in the topic.

Worked example

4 marks

Solve the equation log⁡3(x+5)+log⁡3(x−1)=3\log_3(x + 5) + \log_3(x - 1) = 3.

Show full working
  1. 1

    Two logs added means one log of a product: log⁡3[(x+5)(x−1)]=3\log_3\left[(x+5)(x-1)\right] = 3

    “Combine logarithms into a single logarithm” is a recorded technique — it is nearly always step one.

  2. 2

    Undo the logarithm by writing it as an index statement: (x+5)(x−1)=33=27(x+5)(x-1) = 3^3 = 27

  3. 3

    Expand and collect: x2+4x−5=27x^2 + 4x - 5 = 27, so x2+4x−32=0x^2 + 4x - 32 = 0

  4. 4

    Factorise: (x+8)(x−4)=0(x+8)(x-4) = 0, giving x=−8x = -8 or x=4x = 4.

  5. 5

    Reject x=−8x = -8: it makes log⁡3(x−1)=log⁡3(−9)\log_3(x-1) = \log_3(-9), and you cannot take the log of a negative number.

    Every log equation of this shape needs this check, and the rejection itself is creditworthy.

Answer

x=4x = 4

The domain check is not optional bookkeeping — it is usually the difference between full marks and most of them. Always test both roots in the original logarithms.

Practise the laws of logarithmsReal past-paper questions · Laws of logarithms and relationship with indices

The rest of this note

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Can you do all of these?

  • Rewrite 2x=52^x = 5 as a logarithm and evaluate xx

  • Combine log⁡3(x+5)+log⁡3(x−1)\log_3(x+5) + \log_3(x-1) into a single logarithm

  • Say why log⁡(m+n)\log(m+n) cannot be split up

  • Sketch y=exy = \mathrm{e}^x and y=ln⁡xy = \ln x on one diagram with y=xy = x

  • Describe the difference between y=e2xy = \mathrm{e}^{2x} and y=e−2xy = \mathrm{e}^{-2x}

  • Solve 3 x+1=4 2x−13^{\,x+1} = 4^{\,2x-1} to 3 significant figures

  • Solve e2x−5ex+6=0\mathrm{e}^{2x} - 5\mathrm{e}^{x} + 6 = 0 by substitution

  • Decide whether to plot ln⁡y\ln y against xx or against ln⁡x\ln x for a given law

  • Recover kk from an intercept of ln⁡k\ln k

Now do the questions
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