The laws of logarithms

For a base $a$ with $a > 0$ and $a \neq 1$, and for positive numbers $x$ and $y$,

There is also the change-of-base rule, which turns a logarithm in one base into a quotient of logarithms in another:

Before you start

You need three things:

• What a logarithm is. $\log_a x$ is the power you raise $a$ to in order to get $x$. So $\log_a x = y$ means exactly $a^y = x$.
• The laws of indices. $a^m \times a^n = a^{m+n}$, $a^m \div a^n = a^{m-n}$, and $(a^m)^n = a^{mn}$.
• The rules for the base. A base must satisfy $a > 0$ and $a \neq 1$. The proof shows where each rule is needed.

Predict first

The first law says

The left-hand side multiplies the two inputs. The right-hand side adds the two logarithms.

Before you read on, predict: a logarithm is an index, the power you raise the base to. Multiplying two powers of the same base adds their indices, so why should multiplying $x$ and $y$ add their logarithms? Write your answer in one sentence. Then read the proof and compare.

The proof

One idea runs through all three laws: a logarithm is an index. So each law is an index law read backwards.

Write the two logarithms as letters:

By the definition of a logarithm, that means

Now take the three index laws in turn.

The product law. Multiply $x$ and $y$:

The first index law added the indices. Read the result as a logarithm: the power of $a$ that gives $xy$ is $u + v$. So

The quotient law. Divide $x$ by $y$ instead:

The second index law subtracted the indices, so

The power law. Raise $x$ to the power $k$:

The third index law multiplied the indices, so the power that gives $x^k$ is $uk$:

That is all three laws. Each one is an index law, with the logarithm read off at the end, and none of them is a new fact about logarithms.

Two values worth keeping. Because $a^1 = a$, we have $\log_a a = 1$. Because $a^0 = 1$, we have $\log_a 1 = 0$. The second is why $\log_b a$ is zero only when $a = 1$.

The change-of-base rule. This rule lets you work in a base of your choosing, so it is worth deriving rather than quoting. Let

Take $\log_b$ of both sides:

The power law, already proved, brings the $c$ down:

Now divide both sides by $\log_b a$. This is allowed because $\log_b a$ is not zero: if $\log_b a = 0$ then $a = 1$, and the base $a$ is not $1$. So

What is assumed, and why

Each rule on the base and the inputs is doing a job. The proof would break without it.

• $a > 0$. A logarithm is an index on the base $a$. A negative base has no real power for many exponents — $(-2)^{1/2}$ is not a real number — so a base that works for every input must be positive.
• $a \neq 1$. If $a = 1$ then $1^y = 1$ for every $y$. No power of $1$ gives a number other than $1$, so $\log_1 x$ has no answer when $x \neq 1$, and too many answers when $x = 1$. The base cannot be $1$.
• $x > 0$ and $y > 0$. Since $x = a^u$ and $y = a^v$, and a positive base raised to any power is positive, $x$ and $y$ must be positive. A logarithm of zero or of a negative number has no real value, so the laws are stated for positive inputs only.
• $b > 0$ and $b \neq 1$ in the change-of-base rule, for the same two reasons. The division by $\log_b a$ is the third place the base matters: if $a = 1$ then $\log_b a = 0$ and the fraction has no value.

Check your understanding

1. Write $\log_3 5 + \log_3 7$ as a single logarithm.
2. Use the power law to show that $\log_a\left(\dfrac{1}{x}\right) = -\log_a x$.
3. Explain why $\log_2(xy)$ is defined for $x = -2$ and $y = -3$ when you multiply first, but the law $\log_2 x + \log_2 y$ is not defined for those inputs.

One more thing

The change-of-base rule is why a calculator needs only two logarithm buttons, one for base $10$ and one for base $e$. To work out $\log_2 10$, write

You may use any base for the quotient. The two logarithms change together as the base changes, so the ratio stays the same.