Overview
What you’ll learn
- Sketch and read graphs of y = aˣ (a > 1 and 0 < a < 1), y = eˣ, and y = log_a x / y = ln x.
- Convert between index form y = aˣ and logarithm form x = log_a y.
- Apply the product, quotient and power laws of logarithms to simplify expressions.
- Change the base of a logarithm and use it to evaluate or rearrange.
- Solve exponential equations (including quadratic-in-disguise forms) and logarithmic equations, checking domains.
- Use exponential models to find values and solve for time.
In 30 seconds
- 1y = aˣ has y-intercept (0, 1); a > 1 grows, 0 < a < 1 decays; y = log_a x is its mirror in y = x, crossing at 1 on the x-axis.
- 2Logarithm = exponent: y = aˣ ⟺ x = log_a y. lg x is base 10; ln x is base e.
- 3Product law: log_a (mn) = log_a m + log_a n. Quotient law: log_a (m/n) = log_a m − log_a n.
- 4Power law: log_a (mᵏ) = k log_a m — it pulls the exponent to the front.
- 5Change of base: log_a x = lg x/lg a (any base works, choose the calculator’s).
- 6Solve aˣ = b by taking logs of both sides: x lg a = lg b.
- 7Watch the domain: log arguments must be strictly positive, so log equations need a final check.
- 8Hidden quadratics: 3^(2x) = (3^x)² — substitute y = 3^x, solve, then return to x, rejecting negative y.
Notes · 1 of 5
Exponential functions and their graphs
Multiply by the same factor every step: a curve through (0, 1) that never touches the x-axis.
An exponential function has the variable in the exponent: y = aˣ with a > 0, a ≠ 1. Doubling money, growing bacteria and decaying samples all follow this shape.
| Feature | y = aˣ, a > 1 | y = aˣ, 0 < a < 1 |
|---|---|---|
| y-intercept | (0, 1) | (0, 1) |
| Direction | increasing (growth) | decreasing (decay) |
| Left end | approaches 0 from above | rises steeply |
| x-axis | asymptote — never touched | asymptote — never touched |
The natural exponential y = eˣ (e ≈ 2.718) is the base-a > 1 case used across calculus next year. Reading points off these graphs is examinable: for y = 2ˣ, the point x = 3 gives y = 8, and x = −1 gives y = 1/2.
Notes · 2 of 5
Logarithms and their laws
A logarithm is an exponent in disguise, and the three laws are the index laws re-dressed.
The logarithm log_a x answers: “a to what power gives x?” The two forms are equivalent: y = aˣ ⟺ x = log_a y. So log₂ 8 = 3 because 2³ = 8, and lg 1000 = 3 because 10³ = 1000. Convention here: lg x means log₁₀ x and ln x means logₑ x.
| Law | Statement |
|---|---|
| Product | log_a (mn) = log_a m + log_a n |
| Quotient | log_a (m/n) = log_a m − log_a n |
| Power | log_a (mᵏ) = k log_a m |
The power law pulls exponents out: lg (√5) = ½ lg 5, and ln e³ = 3. It also runs backwards — 2 lg 5 = lg 25 — which is how multi-term expressions collapse. Two special values are worth memorising: lg 10 = 1 and ln e = 1, and logs of 1 are always 0 in any base.
Notes · 3 of 5
Change of base
Any logarithm can be rewritten in the calculator’s base — the ratio of two logs.
Calculators only offer lg and ln, but questions use other bases. The change-of-base formula fixes this: log_a x = lg x/lg a (ln works equally well in both places).
- 1Write the ratio
log_a x = (log of x)/(log of a), both in one chosen base.
- 2Evaluate or simplify
For log₂ 10: lg 10/lg 2 ≈ 3.32.
- 3Exploit exact cases
When both numbers are powers of one base, cancel: log₈ 32 = 5 lg 2/(3 lg 2) = 5/3.
Change of base also rearranges unknowns: expressions like log₇ 27 can be written in terms of a given log₃ 7 by choosing base 3 and spotting the reciprocal structure. Expect one such “express in terms of k” part in harder questions.
Notes · 4 of 5
Solving exponential and logarithmic equations
Take logs to pull x down from the exponent; convert to index form to free x from the log.
Exponential equations (unknown in the exponent) are solved by taking logarithms of both sides. Logarithmic equations (unknown inside a log) are solved by converting to index form — and always checking that the answers keep every log argument positive.
- 1aˣ = b
Take lg of both sides: x lg a = lg b, so x = lg b/lg a.
- 2a^(kx + c) = b
Same method: (kx + c) lg a = lg b, then solve the linear equation.
- 3log_a (expression) = number
Rewrite as a^(number) = expression and solve the result.
- 4Hidden quadratic
For equations in 3^(2x) and 3^x, substitute y = 3^x, solve, then convert back.
Notes · 5 of 5
Exponential and logarithmic models
Growth and decay formulas answer “how much?” directly and “when?” with logs.
An exponential model has the form quantity = initial × (rate)^(time). Questions come in two kinds: evaluate (substitute t, compute) and invert (set the quantity and solve for t with logs).
When the numbers are less friendly, take logarithms of both sides. A decay model M = 80 × 0.9^t reaches half mass when 0.9^t = 0.5, so t = ln 0.5/ln 0.9 ≈ 6.58 — the model’s “half-life” in years. Report model answers with units and to 3 significant figures.
Revise
Key terms
The words examiners expect you to use precisely.
- Exponential function
- A function of the form y = aˣ (a > 0, a ≠ 1), with the variable in the exponent.
- Natural exponential
- The exponential with base e ≈ 2.718, written eˣ or exp x.
- Logarithm
- log_a x is the power to which a must be raised to give x; the inverse of exponentiation.
- Common logarithm
- lg x, the logarithm to base 10.
- Natural logarithm
- ln x, the logarithm to base e.
- Product law
- log_a (mn) = log_a m + log_a n.
Show 6 more key termsShow fewer key terms
- Quotient law
- log_a (m/n) = log_a m − log_a n.
- Power law
- log_a (mᵏ) = k log_a m.
- Change of base
- log_a x = lg x/lg a, valid for any positive base other than 1.
- Asymptote
- A line the graph approaches but never reaches; y = 0 for every y = aˣ graph.
- Growth model
- quantity = initial × (rate)^time with rate > 1, e.g. population doubling.
- Decay model
- quantity = initial × (rate)^time with 0 < rate < 1, e.g. half-life problems.
Revise
Common mistakes
Where students most often lose marks in this chapter — and what to do instead.
Mistake: Believing lg 7 = lg 3 + lg 4.
The product law merges sums of logs, not sums of numbers inside one log: lg 3 + lg 4 = lg 12.
Instead: Logarithms convert multiplication into addition — never addition into addition.
Mistake: Writing log_a (m + n) = log_a m + log_a n.
There is no logarithm law for a sum inside the log.
Instead: Only products, quotients and powers rewrite.
Show 8 more mistakesShow fewer mistakes
Mistake: Forgetting the power law runs both ways: leaving 2 lg 5 unmerged, or “simplifying” lg 4 as 2.
2 lg 5 = lg 25, and lg 4 is a logarithm, not the number 4.
Instead: k log_a m = log_a (mᵏ). lg 4 ≈ 0.602; the exponent only comes out with the power law.
Mistake: Solving aˣ = b by dividing: claiming 3ˣ = 20 gives x = 20/3.
x is an exponent; division does not extract it.
Instead: Take logs of both sides: x = lg 20/lg 3.
Mistake: Taking lg of only one side: lg(3ˣ) = 20.
That changes the equation; both sides must be logged together.
Instead: lg(3ˣ) = lg 20, i.e. x lg 3 = lg 20.
Mistake: Accepting negative y from the substitution, e.g. keeping y = −1/2 when solving for 3ˣ.
3ˣ is positive for every real x, so a negative y cannot happen.
Instead: Reject negative y values before converting back to x.
Mistake: Not checking the domain in log equations, e.g. keeping x = −3 from ln(x + 2) + ln(x − 2) = ln 5.
Log arguments must be strictly positive; x = −3 makes both arguments negative.
Instead: After solving, test every answer in each log argument and reject failures.
Mistake: Reading the y-intercept of y = aˣ as (0, a) or (1, 0).
a⁰ = 1, so the curve always passes through (0, 1).
Instead: Every y = aˣ has y-intercept (0, 1); every y = log_a x crosses the x-axis at (1, 0).
Mistake: Writing log₂ 10 as lg 2/lg 10 (inverted change of base).
The wanted value goes on top: log_a x = lg x/lg a.
Instead: log₂ 10 = lg 10/lg 2 ≈ 3.32; the inverted ratio gives log₁₀ 2 ≈ 0.301.
Mistake: Rounding model answers to 2 significant figures, or dropping units in interpretation.
The syllabus convention is 3 s.f. for non-exact answers, and the interpretation sentence carries a mark.
Instead: t = 6.58 years, V = 22 500 dollars — numbers with units in context.
Check yourself
Quick quiz
18 questions with instant explanations. About 5 minutes.
Question 1 of 3
- Question 1: current question
- Question 2: not answered
- Question 3: not answered
Read exponential graphs
The y-intercept of the graph of y = 3ˣ is
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Practise
Topical questions
Exam-style practice in three tiers, from recall to data questions.
- 5Foundation
- 5Standard
- 5Challenging
15 questions worth 64 marks, each with a model answer, mark scheme and examiner tips.
Practise topical questionsExam practice
Practice paper
A 30-mark paper set out like a school paper, with a separate mark scheme. Give yourself 40 min.
Question paper
Chapter 5 Practice Paper
30 marks · 40 min · 4 pages
5 multiple-choice questions and structured questions with answer lines. Chapter 5 (A6 Exponential and logarithmic functions).
Mark scheme
PROChapter 5 Practice Paper — mark scheme
Answers · Marking points · 2 pages
Model answers with a mark for each creditworthy point — for marking at home or in class.
Where this chapter is tested
Each school sets its own scope — check your assessment outline.
Questions
Frequently asked questions
What does Sec 3 A-Math Chapter 5 cover?
It covers syllabus sub-topic A6 of O-Level 4049 / SEC G3 K341: exponential functions aˣ and eˣ with their graphs, logarithms log_a x and ln x, the three laws of logarithms, change of base, solving exponential and logarithmic equations (including hidden quadratics), and exponential growth and decay models.
How do I solve an equation like 3ˣ = 20?
Take logarithms of both sides: x lg 3 = lg 20, so x = lg 20/lg 3 ≈ 2.73. The log base does not matter as long as both sides use the same one.
Why do log equations sometimes have fake answers?
Log arguments must be strictly positive. Squaring or rearranging can produce candidates that make an argument zero or negative, so each answer must be substituted back into every log argument and invalid ones rejected.
What is the fastest way to evaluate log₈ 32?
Notice both numbers are powers of 2: log₈ 32 = lg 32/lg 8 = 5 lg 2/(3 lg 2) = 5/3. The change-of-base ratio collapses to the ratio of exponents.
When is this chapter usually tested in school?
Exponential and logarithmic functions typically appear in Weighted Assessment 2 (Term 2) and the end-of-year examination. Scope and weighting vary by school, so check your school’s assessment outline.
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Chapter 6: Coordinate Geometry and Circles