mathsGrade 11-122 min read

Logarithms: the laws, and what a log really asks

A logarithm is just the question 'what power?' Once you see that, the laws stop being arbitrary and become a way of turning multiplication into addition — and solving equations with x in the exponent.

Students find logarithms alien because nobody tells them what a log is. It's not a new operation to fear — it's a question. $\log_a b$ asks: "what power do I raise $a$ to, to get $b$ ?" That single sentence makes everything else follow.

The definition, both ways round

\log_a b = c \quad\Longleftrightarrow\quad a^{c} = b

These two lines say exactly the same thing. $\log_2 8 = 3$ because $2^3 = 8$ . A logarithm and an exponential are the same fact read in opposite directions — that's why logs are how you "undo" a power.

The three laws

They look like rules to memorise, but each is an index law in disguise:

\log_a(xy) = \log_a x + \log_a y

\log_a\!\left(\frac{x}{y}\right) = \log_a x - \log_a y

\log_a\!\left(x^{k}\right) = k\,\log_a x

The first one is the heart of it: logs turn multiplication into addition. (That's literally why they were invented — to make hard multiplications into easy sums.) The third is the one that does real work, because it pulls a power down to the front.

Why they matter: x stuck in an exponent

You can't solve $3^x = 20$ with normal algebra — $x$ is trapped up in the power. Logs are the key that releases it. Take a log of both sides and use the third law to bring the $x$ down:

3^x = 20 \;\Rightarrow\; x\log 3 = \log 20 \;\Rightarrow\; x = \frac{\log 20}{\log 3} \approx 2.73

Two values to keep handy

\log_a 1 = 0 \quad (\text{any } a^0 = 1), \qquad \log_a a = 1 \quad (a^1 = a)

You can only combine logs with the same base. $\log_2 5 + \log_3 4$ does not simplify. The laws are about adding powers of the same number — exactly the rule from indices, seen from the other side.

Last revised 25 October 2025.

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