State and apply the product, quotient and power laws of logarithms and the change-of-base relationship to simplify and evaluate expressions

What this dot point is asking

SEAB wants you to use the laws of logarithms to combine several logs into one, split one log into several, and evaluate logarithms by changing the base. A logarithm answers the question "to what power must the base be raised", so $\log_a b = c$ means $a^c = b$. The laws mirror the index laws exactly.

The answer

The meaning of a logarithm

The logarithm is the inverse of an index:

You can only take the logarithm of a positive number, which is why solutions sometimes have to be rejected.

The three laws

For the same base $a$:

A product becomes a sum, a quotient becomes a difference, and a power comes down as a multiplier.

Special values

Two values fall straight out of the definition:

Change of base

To evaluate a logarithm in a base your calculator does not have, change the base to one it does (such as $10$ or $e$):

The common logarithm $\lg x$ means $\log_{10} x$, and the natural logarithm $\ln x$ means $\log_e x$; both are on the calculator, so change-of-base lets you compute any logarithm through one of them.

Combining the laws in one expression

Most questions need the laws together: bring powers down first, then merge products into sums and quotients into differences, working towards a single logarithm or a numerical value. A common target form is $\log_a$ of a single simplified number, from which a value follows at once.

Expressing one logarithm in terms of given ones

A frequent A-Maths task gives you $\log_a 2 = p$ and $\log_a 3 = q$ and asks for the logarithm of some related number. The method is to factorise that number into powers of $2$ and $3$, then apply the laws to break the logarithm into the given pieces. For $\log_a 12$, write $12 = 2^2 \times 3$, so $\log_a 12 = 2\log_a 2 + \log_a 3 = 2p + q$. Even fractions work: $\log_a 1.5 = \log_a \tfrac{3}{2} = q - p$. The skill is the prime factorisation that exposes the given building blocks, after which the log laws do the rest.

Watching the domain when solving log equations

Because a logarithm only accepts a positive argument, every solution to a logarithmic equation must be checked against the domain, and invalid roots discarded. After combining $\log_2 x + \log_2(x - 2) = 3$ into a quadratic with roots $x = 4$ and $x = -2$, only $x = 4$ survives, because $x = -2$ would make both $\log_2 x$ and $\log_2(x - 2)$ undefined. The reliable habit is to state the required domain ($x > 2$ here) before solving, so any candidate outside it is rejected on sight rather than overlooked. This domain check is where method marks are commonly lost.

Examples in context

Example 1. Decibels and pH. Sound level in decibels and acidity in pH are logarithmic scales, so a tenfold change in the underlying quantity is a fixed step on the scale. The product and quotient laws are what let scientists add and subtract these levels.

Example 2. Simplifying before solving. Before solving an equation like $\log x + \log(x - 3) = 1$, the product law combines the left side into a single logarithm, after which the equation converts to a quadratic, the standard route in logarithmic equations.

Try this

Q1. Simplify $\log_5 50 - \log_5 2$. [2 marks]

• Cue. Quotient law: $\log_5 25 = 2$.

Q2. Express $\log_2 24$ in terms of $\log_2 3$. [2 marks]

• Cue. $24 = 2^3 \times 3$, so $\log_2 24 = 3 + \log_2 3$.

Q3. Use change of base to evaluate $\log_4 64$. [2 marks]

• Cue. $\dfrac{\log 64}{\log 4} = \dfrac{3\log 4}{\log 4} = 3$.