How is change of base written?

log2 10 = log10 10 / log10 2 = 1 / 0.3010, about 3.32.

What if the unknown is an exponent?

log8 2 = 1/3 since 8^(1/3) = 2, and log9 3 = 1/2 since 9^(1/2) = 3.

Why check roots of a log equation?

log4 32 = log2 32 / log2 4 = 5 / 2 = 2.50.

S4 Maths: Logarithms Paper 2 (Change of Base and Equations)

2026-11-03

S · 4

This paper focuses on change of base and log equations: matching bases, taking logs, and checking that the numbers stay positive.

Quick answer　1. 3.32 | 2. 1/3 and 1/2 | 3. 2.5 | 4. x = 3 | 5. 5.32 | 6. x = 4 | 7. x = 5 | 8. 3 | 9. 0.793 and 3.585 | 10. x = 7

01　Paper details (check answers afterwards)

Item	Detail
Questions	10 questions
Suggested time	25 minutes
Difficulty mix	Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1
Chapter	Chapter: S4 Maths — Logarithms
Level	Application

02　Questions (10 items)

FoundationUse change of base to find log2 10, taking log10 2 as about 0.3010.
FoundationFind log8 2 and log9 3.
FoundationFind log4 32.
FoundationSolve 3^(x+1) = 81.
IntermediateSolve 2^x = 40, taking log10 2 as about 0.3010 and log10 40 as about 1.6021.
IntermediateSolve log2 x + log2 (x – 2) = 3.
IntermediateSolve log10 x + log10 (x – 3) = 1.
ChallengeFind log2 3 x log3 4 x log4 8.
ChallengeGiven log2 3 is about 1.585, find log2 root 3 and log2 12.
Error redoA student solves log2 (x + 1) = 3 and gets x = 9 by turning 2^3 into 3^2. What is the correct answer?

03　Answers and one-line notes

#	Answer	One-line note
1	log2 10 = log10 10 / log10 2 = 1 / 0.3010, about 3.32.	Change of base turns an awkward base into one the calculator provides.
2	log8 2 = 1/3 since 8^(1/3) = 2, and log9 3 = 1/2 since 9^(1/2) = 3.	When the number is a root of the base, the answer is that simple fraction.
3	log4 32 = log2 32 / log2 4 = 5 / 2 = 2.50.	Both 4 and 32 are powers of 2, so switching to base 2 solves it at once.
4	Since 81 = 3^4, x + 1 = 4 and x = 3.	Once both sides share a base, compare the exponents only.
5	x = log10 40 / log10 2, about 1.6021 / 0.3010 = 5.32.	When the unknown sits in the exponent, take logs of both sides to bring it down.
6	x(x – 2) = 8 gives x^2 – 2x – 8 = 0, so (x – 4)(x + 2) = 0; as x > 2, x = 4.	Always check the numbers stay positive; x = -2 makes the logs undefined.
7	x(x – 3) = 10 gives x^2 – 3x – 10 = 0, so (x – 5)(x + 2) = 0; take x = 5.	Of the two roots only x = 5 keeps both numbers positive.
8	Chaining the change of base leaves log2 8 = 3.	Neighbouring base and number cancel, leaving only the first and last.
9	log2 root 3 = 1.585 / 2 = 0.792, and log2 12 = 2 + 1.585 = 3.585.	A square root halves the log, and 12 splits into 4 x 3.
10	x + 1 = 2^3 = 8, so x = 7.	In index form the base stays put; only the exponent moves across.

04　What to do after marking

After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
Repeat the same item two days later; move to the next paper only when both attempts are correct.
Parents only need the answer and note columns; there is no need to explain every item.

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