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Logarithms and the log laws

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What a logarithm is

A logarithm answers the question "what power?"

$\log_a b = c \quad \text{means} \quad a^c = b$

Each part:
- $a$ — the base, which must be positive and not 1
- $b$ — the number you are taking the log of, which must be positive
- $c$ — the answer: the power the base must be raised to
$\log_2 8 = 3$ because $2^3 = 8$ .
$\log_{10} 1000 = 3$ because $10^3 = 1000$ .
Logarithms appear in this standard for one reason only: they are the tool that solves exponential equations, where the unknown is in the power.

The two forms

Index form and log form say the same thing:

$a^c = b \quad \Longleftrightarrow \quad \log_a b = c$

Converting between them is the core skill. Practise it in both directions:
- $5^2 = 25$ becomes $\log_5 25 = 2$
- $\log_3 81 = 4$ becomes $3^4 = 81$

Common and natural logarithms

$\log x$ with no base written means $\log_{10} x$ — the common logarithm.
$\ln x$ means $\log_e x$ — the natural logarithm, where $e \approx 2.718$ .
Your calculator has both, and either can be used to solve an exponential equation.

Values worth knowing

$\log_a 1 = 0$ for every base, because $a^0 = 1$ .
$\log_a a = 1$ , because $a^1 = a$ .
$\log_a\left(a^n\right) = n$ — the log and the power undo each other.
You cannot take the log of zero or of a negative number. Any solution requiring one must be rejected.

The log laws

Three rules, each mirroring an index law:

$\log_a(MN) = \log_a M + \log_a N$

$\log_a\left(\frac{M}{N}\right) = \log_a M - \log_a N$

$\log_a\left(M^n\right) = n\log_a M$

Multiplication becomes addition. Division becomes subtraction. A power comes down to the front.
These are the index laws in disguise: $a^m \times a^n = a^{m+n}$ is exactly the first rule read backwards.
The third law is the one that matters most, because bringing the power down is what frees an unknown out of an exponent.

The rules that do not exist

$\log(M + N)$ is not $\log M + \log N$ . There is no law for the log of a sum.
$\dfrac{\log M}{\log N}$ is not $\log\left(\dfrac{M}{N}\right)$ . The subtraction law needs the division inside the log.
$\left(\log M\right)^2$ is not $2\log M$ . The power must be inside the log for the third law to apply.

Combining and splitting

Splitting up — useful before differentiating or simplifying:

$\log\left(\frac{x^3y}{z}\right) = 3\log x + \log y - \log z$

Combining into one log — essential before solving an equation:

$2\log x + \log 3 = \log\left(3x^2\right)$

To solve a log equation, get a single log on each side, then equate what is inside:

$\log A = \log B \quad \Rightarrow \quad A = B$

Worked through — solve $\log_2 x + \log_2(x - 2) = 3$ $lo g_{2} x + lo g_{2} (x - 2) = 3$ :
- Combine: $\log_2\left(x(x-2)\right) = 3$
- Convert to index form: $x(x-2) = 2^3 = 8$
- Solve: $x^2 - 2x - 8 = 0$ , so $(x-4)(x+2) = 0$ , giving $x = 4$ or $x = -2$
- Reject $x = -2$ , because $\log_2(-2)$ does not exist
- $x = 4$ is the only solution

Tips and Tricks

Always check log solutions against the domain. A log needs a positive argument, so any root that makes something inside a log zero or negative must be rejected with a reason. This is the single most common way to lose a mark on log equations.

$\log(M+N)$ cannot be split. There is no law for the log of a sum, and inventing one is a standard error. The 2025 report lists "failed to manipulate expressions that involved surds, indices, or logs" among the Not Achieved behaviours.

Bring the power down first. $\log\left(x^5\right) = 5\log x$ is what makes an unknown accessible. If your unknown is still stuck in an exponent, you have not used the third law yet.

Combine to a single log before converting to index form. $\log A + \log B = c$ must become $\log(AB) = c$ before it becomes $AB = a^c$ .

Converting between log form and index form is worth practising until it is instant. Most log questions are really index questions wearing a disguise.

The report lists "solved log equations" as an Achieved behaviour and "manipulated logs" as Merit — so the laws are expected, not optional.

Worked ExampleSolving a logarithmic equation

Solve $\log_5(x + 3) + \log_5(x - 1) = 1$ .

Step 1 — Note the domain restrictions before starting

Both arguments must be positive:

$x + 3 > 0 \;\Rightarrow\; x > -3$
$x - 1 > 0 \;\Rightarrow\; x > 1$

The stricter condition wins, so any valid solution must satisfy $x > 1$ .

Step 2 — Combine the two logs into one

Both logs have the same base, and they are added, so use $\log M + \log N = \log(MN)$ :

$\log_5\left[(x+3)(x-1)\right] = 1$

Step 3 — Convert to index form

$\log_a b = c$ means $a^c = b$ . Here the base is 5 and the answer is 1:

$(x+3)(x-1) = 5^1 = 5$

Step 4 — Expand and rearrange

$x^2 - x + 3x - 3 = 5$

$x^2 + 2x - 3 = 5$

$x^2 + 2x - 8 = 0$

Step 5 — Solve the quadratic

Two numbers multiplying to $-8$ and adding to $2$ are $4$ and $-2$ :

$(x + 4)(x - 2) = 0$

$x = -4 \quad \text{or} \quad x = 2$

Step 6 — Test each against the domain

$x = -4$ : substituting into the original gives $\log_5(-1) + \log_5(-5)$ . Logs of negative numbers do not exist, and $-4$ also fails the condition $x > 1$ from Step 1.

$x = 2$ : gives $\log_5(5) + \log_5(1)$ . Both arguments are positive ✓ and $2 > 1$ ✓

Step 7 — Verify the surviving solution

$\log_5(2+3) + \log_5(2-1) = \log_5 5 + \log_5 1 = 1 + 0 = 1 \;\checkmark$

Note that the rejected root was a perfectly good solution of the quadratic — it was created by the algebra, not by the original equation. That is why the domain check is compulsory, not optional.