A-Maths Logarithms: Check the Domain Before Solving

Give each logarithm a meaning

In a public SGExams maths-help thread, a Secondary 3 student identified logarithms as their weakest topic and asked how to improve beyond doing more questions. A useful response is to make every transformation explainable: what power does the logarithm represent, which law applies, and which values are allowed?

The official 2026 Additional Mathematics syllabus includes logarithm laws, conversion between exponential and logarithmic forms, change of base, and solving simple equations. The examples below are original practice illustrations.

The statement log28 = 3 means 23 = 8. The logarithm is the exponent. Read several examples in both directions before manipulating longer expressions. For real logarithms, the argument must be positive; the base must be positive and different from one. Those conditions belong to the meaning of the expression.

Write the domain before combining terms

Consider log2(x − 1) + log2(x + 2) = log218. The first logarithm requires x − 1 > 0, and the second requires x + 2 > 0. Together, these conditions give x > 1. Write that restriction beside the equation so that it remains visible.

The logarithms on the left have the same base, so the product law gives log2[(x − 1)(x + 2)] = log218. Equal logarithms with the same valid base have equal arguments. Therefore (x − 1)(x + 2) = 18.

Expanding and rearranging gives x² + x − 20 = 0, so (x + 5)(x − 4) = 0. The candidate roots are x = −5 and x = 4. Only x = 4 satisfies x > 1. Substituting it into the original equation gives log23 + log26 = log218, as required.

The rejected root illustrates an important point. At x = −5, the product (x − 1)(x + 2) is positive, but each original argument is negative. The combined expression can be defined at a value where the original sum is not. Keeping the original domain prevents that extra value from becoming an answer.

Separate three different operations

The product law says logau + logav = loga(uv), for positive u and v. It does not say that the sum equals loga(u + v). To see the distinction, log22 + log24 equals 1 + 2 = 3, whereas log26 is not 3.

The power law moves an exponent on an argument: loga(uk) = k logau when u is positive. Changing a base is a separate operation. For example, log864 = log264 ÷ log28 = 6 ÷ 3 = 2. Simply replacing the base 8 with 2 would change the value.

A calculator can help you notice a mistaken rule, but agreement in one example does not establish a general identity. After a numerical check, return to the law and its conditions. This is especially useful when two expressions happen to match for a convenient input.

Practise the decision before the calculation

For a short revision session, take three questions and write only the domain and the proposed first step. Decide whether to combine logarithms, convert to exponential form, or change the base. Explain why that step makes the equation simpler before completing the algebra.

When reviewing an error, classify it precisely. Was the argument read incorrectly? Were unlike bases combined? Was a sum placed inside a logarithm? Was a candidate root accepted without checking the original expression? These lead to different repairs, even if they produce the same wrong final answer.

Use the TestPapersFree Secondary A-Maths collection for suitable follow-up questions. Start with a small selection on logarithms, then include one in mixed practice. Keep the domain line and the final substitution check even when the middle steps become familiar. They explain why the answer belongs to the original problem.