A logarithm answers a single question: what power do I need to raise a base to, to get a given number? log₂ 8 = 3 simply means '2 to the power of 3 equals 8'. Every rule in this topic follows from that one idea, and most of the errors students make come from losing sight of it and treating logs as an unfamiliar symbol to manipulate blindly.
The definition, and moving between forms
For example, log₃ 81 = 4 because 3⁴ = 81. When an equation has an unknown trapped inside a logarithm or inside an exponent, converting between these two forms is usually the first move that unlocks the question.
The three laws of logarithms
| Law | Statement | Common use |
|---|---|---|
| Addition law | logₐ x + logₐ y = logₐ (xy) | Combining separate logs into one |
| Subtraction law | logₐ x − logₐ y = logₐ (x/y) | Combining a difference of logs into one |
| Power law | n logₐ x = logₐ (xⁿ) | Moving a power out of or into a log |
Two further results are used constantly: logₐ 1 = 0 for any base a (since a⁰ = 1), and logₐ a = 1 for any base a (since a¹ = a). Both are worth knowing automatically rather than deriving each time.
Worked example: combining logs
Write log 5 + 2 log 3 − log 45 as a single logarithm.
- Apply the power law to 2 log 3: this becomes log 9.
- The expression is now log 5 + log 9 − log 45.
- Apply the addition law to the first two terms: log 5 + log 9 = log 45.
- The expression becomes log 45 − log 45 = log 1 = 0.
Solving exponential equations
An exponential equation has the unknown in the power, such as 5ˣ = 30. Since 30 is not a neat power of 5, you cannot solve this by inspection — you take logs of both sides.
- Take log (or ln) of both sides: log(5ˣ) = log(30).
- Apply the power law to bring the exponent down: x log 5 = log 30.
- Divide: x = log 30 ÷ log 5 ≈ 2.113.
This method works with any consistent base of logarithm — base 10 (log) and base e (ln) both give the same final answer, since the base cancels out in the division. Choose whichever your calculator handles most directly.
Worked example: an equation with the unknown appearing twice
Solve 3²ˣ − 10(3ˣ) + 9 = 0.
- Notice 3²ˣ = (3ˣ)², so this is a quadratic in disguise. Let y = 3ˣ.
- The equation becomes y² − 10y + 9 = 0.
- Factorise: (y − 1)(y − 9) = 0, so y = 1 or y = 9.
- Substitute back: 3ˣ = 1 gives x = 0 (since 3⁰ = 1); 3ˣ = 9 gives x = 2 (since 3² = 9).
The natural logarithm and e
e is a specific irrational number, approximately 2.71828, that arises naturally in continuous growth and decay contexts, and ln x means logₑ x. All three laws of logarithms apply to ln exactly as they do to log base 10 — only the base changes.
Two results are used repeatedly in A-Level Pure and applied questions: eˡⁿˣ = x and ln(eˣ) = x, because ln and the exponential function eˣ undo each other completely (they are inverse functions). This is what makes ln the natural choice when solving equations that already involve e.
Worked example: solving with e
Solve e²ˣ⁺¹ = 20.
- Take ln of both sides: ln(e²ˣ⁺¹) = ln 20.
- Since ln and e cancel: 2x + 1 = ln 20.
- Rearrange: x = (ln 20 − 1) ÷ 2 ≈ 0.998.
Logarithmic equations and the domain restriction
Logarithms are only defined for positive arguments — you cannot take the log of zero or a negative number. This matters directly when solving equations, because algebra can produce a solution that is not actually valid.
Worked example: rejecting an invalid solution
Solve log₂(x + 3) + log₂(x − 1) = 5.
- Combine using the addition law: log₂[(x + 3)(x − 1)] = 5.
- Convert to exponential form: (x + 3)(x − 1) = 2⁵ = 32.
- Expand and rearrange: x² + 2x − 3 = 32, so x² + 2x − 35 = 0.
- Factorise: (x + 7)(x − 5) = 0, so x = −7 or x = 5.
- Check validity: x = −7 makes x − 1 = −8, and you cannot take log₂ of a negative number, so this solution is rejected. x = 5 gives x + 3 = 8 and x − 1 = 4, both positive, so x = 5 is the only valid solution.
Common errors summary
| Error | Correct approach |
|---|---|
| logₐ x + logₐ y = logₐ (x + y) | Should be logₐ (xy) — values multiply, not add |
| Not checking log arguments stay positive | Reject any root that makes an argument zero or negative |
| Treating log(x + y) as log x + log y | There is no law for the log of a sum — it cannot be split |
| Losing track of which base a calculator button uses | Confirm log means base 10 and ln means base e on your calculator |
| Not spotting a disguised quadratic in aˣ | Substitute y = aˣ whenever a²ˣ term and aˣ term both appear |
How to revise this topic
- Memorise the three laws in words, not just symbols, so you can explain out loud what each one does.
- Practise switching between exponential and log form until it's automatic — most equations start by requiring that switch.
- Always check the domain on any question combining logs with a quadratic.
- Practise a mix of log-base-10, natural log, and general base a questions, since exam papers move between them without warning.
Logarithms and exponentials sit at the centre of A-Level Pure — they reappear inside differentiation, integration, and modelling questions throughout Year 12 and 13. Getting the three laws and the domain restriction genuinely secure now pays off across the rest of the course, not just in this topic's own questions.
Worked example: change of base
Calculators do not always have a button for an arbitrary base — most only give log (base 10) and ln (base e) directly. The change of base formula lets you evaluate any logₐ x using whichever base your calculator provides.
For example, to evaluate log₇ 50, calculate log 50 ÷ log 7 ≈ 2.0115. This also underlies why solving 5ˣ = 30 by 'taking logs of both sides' works regardless of which base of log you choose — the base cancels between numerator and denominator once you rearrange for x.
Worked example: modelling with exponentials and logs
A population of bacteria is modelled by P = P₀e^(kt), where P₀ is the initial population, t is time in hours, and k is a constant. If the population doubles every 5 hours, find the value of k.
- Doubling means P = 2P₀ when t = 5: 2P₀ = P₀e^(5k).
- Divide both sides by P₀ (it cancels, since it's non-zero): 2 = e^(5k).
- Take ln of both sides: ln 2 = 5k.
- Solve: k = (ln 2) / 5 ≈ 0.1386.
Sketching logarithmic and exponential graphs
Questions frequently ask you to sketch y = logₐ x or y = aˣ and mark key features, without doing any further calculation. Knowing the shape and key points cold saves time and avoids losing marks on features that don't require any working.
| Graph | Key features |
|---|---|
| y = aˣ (a > 1) | Passes through (0, 1); y > 0 for all x; horizontal asymptote y = 0 as x → −∞; increasing |
| y = logₐ x (a > 1) | Passes through (1, 0); defined only for x > 0; vertical asymptote at x = 0; increasing |
y = logₐ x is the reflection of y = aˣ in the line y = x, because logarithm and exponential functions (with the same base) are inverses of each other — this is exactly why one passes through (0, 1) and the other through (1, 0), with the roles of the asymptote and the domain swapped.
A short glossary
- Base: the number being raised to a power — the 'a' in logₐ x or aˣ.
- Argument: the value inside the logarithm — the 'x' in logₐ x. This must always be positive.
- Natural logarithm (ln): logarithm with base e, used throughout calculus and growth/decay modelling.
- Common logarithm (log): logarithm with base 10, the default button on most calculators.
- Inverse functions: two functions that 'undo' each other, such as aˣ and logₐ x with matching bases.