HKDSE Maths | Exponential and Log Functions Tutor Prep Guide | Base change, equations, growth and decay

2026-11-01

Tutor lesson-prep guides · Secondary Maths

Students can recite the definition of a logarithm, yet freeze at 2^(3x-1) = 16 because they have not built the reflex of matching bases. This lesson drills three moves: unify the base, turn an index equation into a linear one, and turn a word problem into one power expression.

Objectives — Students can: (1) simplify products, quotients and powers using index laws; (2) convert freely between index and log form; (3) solve index equations by matching bases or by taking logs; (4) model growth and decay with a(1+r)^t and solve for the time.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Index laws	What is a^m times a^n	a^(m+n)
Negative and fractional indices	2^-3 and 9^(1/2)	1/8 and 3
Solving linear equations	Solve 3x – 1 = 4	x = 5/3

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-10 min	Index law warm-up	Add indices only when bases match	Simplify a^3 x a^5 / a^2
10-24 min	Definition of a log	log_a b = c means a^c = b	log_3 81 = 4
24-40 min	Solving index equations	Match the base, then compare indices	2^(3x-1) = 2^4
40-52 min	Base change on the calculator	log_a b = log b / log a	log_7 50 is about 2.01
52-60 min	Modelling and exit check	A = P(1+r)^t	10,000 grows to 16,289

03 Teaching script (read it out)

Scene 1: Act 1: index laws must be automatic

Start from a^m x a^n = a^(m+n) and push on to (a^m)^n = a^(mn) and a^0 = 1. Ask students why a^0 = 1: a^3 / a^3 = a^(3-3) = a^0, and a number divided by itself is 1.
Ask: what is 2^5 x 2^-2? (2^3 = 8.)

Task: simplify (3x^2)^3 / (9x). (3x^5.)

Scene 2: Act 2: a logarithm is just another way to write an index

log_a b = c and a^c = b say exactly the same thing. So log_3 81 = 4 because 3^4 = 81, and log_2 32 = 5 because 2^5 = 32. Have students convert both directions ten times until it is automatic. Only then introduce the addition law log(MN) = log M + log N, justified by adding indices.
Ask: what is log_5 1 and why? (0, because 5^0 = 1.)

Task: rewrite log_2 32 = 5 in index form, then rewrite 10^3 = 1000 in log form.

Scene 3: Act 3: equations and applications both run on matching bases

For 2^(3x-1) = 16, write 16 = 2^4, so 3x – 1 = 4 and x = 1.667, that is 5/3. When bases cannot match, as in 3^x = 20, take logs on both sides: x = log 20 / log 3, about 2.73. Application: 10,000 at 5% for ten years gives 10,000 x 1.05^10 = 16,289; doubling takes log 2 / log 1.05, about 14.2 years.
Ask: why is doubling time independent of the principal? (It cancels on both sides.)

Task: solve 3^x = 20 (about 2.73) and explain why matching bases fails here.

04 Question types and worked steps

Type	Example	Steps	Answer
Simplify indices	Products, quotients and brackets	Add indices, handle coefficients separately	3x^5
Log conversion	Index form or log form	log_a b = c means a^c = b	log_3 81 = 4
Solve index equation	Bases can be matched	Unify the base and compare indices	x = 1.667
Growth and decay	Principal, rate, time	A = P(1+r)^t, then take logs	14.2 years

Sanity check (teach this): Always substitute back: 2^(3 x 5/3 – 1) = 2^4 = 16, so it checks. Keep the log base consistent; log 20 / log 3 is not log(20/3). The amount 16,289 must exceed the principal 10,000; a smaller answer means the rate sign is wrong. For decay use (1 – r) with r as a decimal: 5% is 0.05, not 5. Logs are defined only for positive arguments, so log(-3) has no value and should be flagged.

05 Independent practice (3 questions)

Question	Answer	What to watch
Solve 3^x = 81	x = 4	81 = 3^4
Solve 2^(3x-1) = 16	x = 1.667	Match the base first
10,000 at 5% for ten years	16,289	P(1+r)^t

06 Common mistakes and what to say

Mistake	Student writes	Say this
Adding indices of different bases	Writes 2^3 x 2^4 = 2^12	Say: “Same base means add the indices, so 2^3 x 2^4 = 2^7.”
Treating a log quotient as a log of a quotient	Writes log 20 / log 3 = log(20/3)	Say: “Two logs divided cannot be merged; press the calculator twice.”
Not substituting back	Stops at x = 5/3	Say: “Put it back: 2^(3 x 5/3 – 1) must equal 16 before you stop.”
Percent not converted	Writes A = P(1 + 5)^t	Say: “5% is 0.05, not 5; writing 5 means five hundred percent.”

07 Homework (paper, section and question numbers)

Secondary Maths chapter index — Exponential and log set, questions 3 and 6
Prep guide directory — Other chapters in this subject
HKDSE Maths practice papers — Index and log questions 1-4

08 Exit check (last 5 minutes)

Question	Answer	Decision
Value of log_2 32	5	All three right: move to applications
Solve 2^(3x-1) = 16	1.667	One wrong: reteach matching bases
Value of 10,000 at 5% after ten years	about 16,289	Two or more wrong: restart at Act 1

09 Tutor FAQ

Why can logs handle very large numbers?

A log turns multiplication into addition and an exponent into a coefficient, so quantities of very different size fit on one scale.

What is the difference between log and ln?

The base: log in HKDSE usually means base 10, while ln means base e, about 2.718. The base-change formula converts between them.

When must I take logs?

When both sides cannot be written as powers of the same base, for example 3^x = 20.

Find a tutor — free Tutor Registration →

A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.