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Additional Mathematics 06062.1

Laws of logarithms

What a logarithm is, and how to use the laws to solve equations.

Learning objectives

What you need to be able to do

Teacher-mapped phrasing — check against the official Cambridge syllabus for exact wording.

  • 2.1.1Use the laws of logarithms and solve equations involving exponentials and logarithms.

6 minute read

Laws of logarithms

What a logarithm is

A logarithm answers the question "what power?". log_a b = c means exactly aᶜ = b. So log₂ 8 = 3 because 2³ = 8. Every logarithm statement can be rewritten as an index statement, and switching between the two forms is usually the key move.

The laws

These mirror the index laws, because logarithms are indices:

  • log a + log b = log(ab) — addition becomes multiplication.
  • log a − log b = log(a/b) — subtraction becomes division.
  • log(aⁿ) = n log a — a power comes down to the front.
  • log_a a = 1 and log_a 1 = 0.

The third law is the workhorse: it is how an unknown trapped in an exponent gets freed.

Solving exponential equations

To solve 3ˣ = 20, take logs of both sides: x log 3 = log 20, so x = log 20 / log 3 ≈ 2.727. The whole method is: take logs, bring the power down, divide.

Natural logarithms

ln x means log_e x, where e ≈ 2.718. The same laws apply. ln e = 1, and e^(ln x) = x. Exponential growth and decay use y = Ae^(kx), with k positive for growth and negative for decay.

The domain restriction

You cannot take the logarithm of zero or a negative number. When solving, always check your solutions against this — a perfectly valid-looking algebraic answer may have to be rejected.

Think of it like this

A logarithm is the question mark in "2 to the what makes 8?". Because it is an index in disguise, every log law is just an index law read backwards — which is why they are worth learning as pairs.

Worked examples

Method, step by step

Solve 5ˣ = 40, giving your answer to 3 significant figures.

  1. 1Take logarithms of both sides: log(5ˣ) = log 40.
  2. 2Use the power law to bring x down: x log 5 = log 40.
  3. 3Divide: x = log 40 / log 5.
  4. 4x ≈ 1.60206 / 0.69897 ≈ 2.2920.

x ≈ 2.29

Solve log₂ x + log₂ (x − 2) = 3.

  1. 1Combine using the addition law: log₂ (x(x − 2)) = 3.
  2. 2Rewrite in index form: x(x − 2) = 2³ = 8.
  3. 3Expand and rearrange: x² − 2x − 8 = 0, which factorises to (x − 4)(x + 2) = 0.
  4. 4So x = 4 or x = −2. But x = −2 makes log₂ x undefined, so it must be rejected.

x = 4 only

Common misconceptions

  • Writing `log(a + b) = log a + log b`. It is the other way round: **log a + log b = log(ab)**.
  • Forgetting to reject solutions that would require the log of a negative number or zero.
  • Thinking `log(a/b) = log a / log b`. Division inside becomes **subtraction** outside.

In the exam

  • When the unknown is in an exponent, taking logs of both sides is almost always the intended first step.
  • After solving any equation containing logs, substitute your answers back to check none makes a logarithm undefined.