Secondary 3 Additional Mathematics | Singapore G3
Exponential & Logarithmic Functions

Why a Logarithm Is Really Asking “What Power?”

log₂8 = 3 is not a mysterious new operation. It is another way of writing 2³ = 8.

That single equivalence is the centre of the topic.

Students often meet logarithms as a page of laws:

• log(ab) = log a + log b;
• log(a/b) = log a − log b;
• log(aⁿ) = n log a.

If those rules are memorised before the meaning is secure, the topic becomes fragile. Change the base, move a term or add an unfamiliar expression and the student may no longer know what is legal.

The stronger model is: exponentials build outputs from powers; logarithms undo that process by asking which exponent produced the output.

The current Singapore G3 Additional Mathematics syllabus includes exponential and logarithmic functions aˣ, eˣ, log_ax and ln x; their graphs; logarithm laws; the equivalence between exponential and logarithmic form; change of base; simplifying expressions; solving simple equations; and mathematical modelling.

SEAB 2027 G3 Additional Mathematics syllabus (K341) →

School sequencing can differ. This article is written for the common Secondary 3 learning stage, while the official syllabus specifies the subject as a whole rather than a compulsory Sec 3/Sec 4 split.

The Topic Job

Move fluently between repeated multiplicative growth and the exponent required to produce that growth.

The topic has two complementary objects:

• Exponential function: the variable is in the exponent.
• Logarithmic function: returns the exponent needed to obtain a positive input.

Once those objects are understood as inverses, the graphs, equations and laws connect naturally.

Start With Powers You Already Know

Consider powers of 2:

Exponent	Power
−2	1/4
−1	1/2
0	1
1	2
2	4
3	8
4	16

The exponential question is:

If the exponent is 3, what value does 2ˣ produce?

The logarithmic question reverses it:

What exponent must 2 be raised to in order to produce 8?

Answer: 3.

Therefore:

2³ = 8 ⇔ log₂8 = 3.

The Fundamental Equivalence

y = aˣ ⇔ x = log_ay

with the real-logarithm conditions:

• a > 0;
• a ≠ 1;
• y > 0.

These restrictions are not decorative.

For a positive base a, aˣ is always positive, so its inverse logarithm only receives positive inputs.

Why the Base Cannot Be 1

If a = 1:

1ˣ = 1

for every real x.

The function never produces 2, 5 or 1/3, and infinitely many exponents all produce the same output 1. It therefore cannot have a logarithmic inverse function in the ordinary real sense.

Why Logarithm Inputs Must Be Positive

For a > 0:

aˣ > 0

for every real x.

There is therefore no real x satisfying:

2ˣ = −5.

So log₂(−5) has no real value.

A logarithm is not allowed to receive zero or a negative real argument.

The Exponential Graph

For y = aˣ with a > 1:

• domain: all real x;
• range: y > 0;
• passes through (0, 1);
• increases as x increases;
• approaches y = 0 as x → −∞ but never reaches it.

For 0 < a < 1, the graph decreases rather than increases.

Example:

(1/2)ˣ = 2⁻ˣ.

This is exponential decay.

The Logarithmic Graph Is the Inverse Picture

For y = log_ax with a > 1:

• domain: x > 0;
• range: all real y;
• passes through (1, 0);
• increases slowly;
• has vertical asymptote x = 0.

Because exponential and logarithmic functions are inverses, their graphs are reflections in the line y = x.

Coordinates swap:

(0,1) on y = aˣ ↔ (1,0) on y = log_ax.

Why log_a1 = 0

Ask the meaning:

log_a1 = ?

What power of a gives 1?

Since a⁰ = 1:

log_a1 = 0.

Why log_aa = 1

What power of a gives a?

Since a¹ = a:

log_aa = 1.

These should not be isolated facts. They follow immediately from the definition.

The Product Law Comes From Adding Exponents

Suppose:

x = log_aM and y = log_aN.

Then:

aˣ = M and aʸ = N.

Multiply:

MN = aˣaʸ = aˣ⁺ʸ.

Therefore:

log_a(MN) = log_aM + log_aN.

The logarithm law is exponent law translated through the inverse function.

The Quotient Law

log_a(M/N) = log_aM − log_aN.

This comes from:

aˣ/aʸ = aˣ⁻ʸ.

The Power Law

log_a(Mⁿ) = n log_aM.

Again, this is the exponent law:

(aˣ)ⁿ = aⁿˣ

translated into logarithmic language.

The Law That Does Not Exist

There is no general law:

log(a + b) = log a + log b.

Counterexample in base 10:

log(10 + 100) = log 110,
but log 10 + log 100 = 1 + 2 = 3.

The laws correspond to multiplication, division and powers—not addition inside a logarithm.

Worked Example 1 — Simplify Without a Calculator

Simplify:

log₂32 + log₂(1/4).

Use meaning:

log₂32 = 5,
log₂(1/4) = −2.

So the result is:

3.

Or combine first:

log₂(32 × 1/4) = log₂8 = 3.

Two routes agree.

Natural Logarithms and the Number e

The number e ≈ 2.71828… is a special exponential base.

The natural exponential function is:

y = eˣ.

Its inverse logarithm is written:

ln x = log_ex.

At this level, you should know how to use eˣ and ln x algebraically and graphically. Later calculus gives a deeper reason e is so important: eˣ has a uniquely simple relationship with its own derivative.

Change of Base

Your calculator may not have a button for log₇ directly.

The change-of-base formula is:

log_ab = log_cb / log_ca.

Common choices are c = 10 or c = e:

log₇20 = log 20 / log 7 = ln 20 / ln 7.

Where Change of Base Comes From

Let:

x = log_ab.

Then:

aˣ = b.

Take logarithm base c:

log_c(aˣ) = log_cb.

Use the power law:

x log_ca = log_cb.

Therefore:

x = log_cb / log_ca.

Again, the formula can be rebuilt if the meaning is understood.

Solving Exponential Equations With a Common Base

Example:

3²ˣ⁻¹ = 27.

Rewrite 27 as 3³:

3²ˣ⁻¹ = 3³.

Because the base is the same and 3ˣ is one-to-one:

2x − 1 = 3 → x = 2.

This method is best when both sides can be expressed naturally with the same base.

Solving an Exponential Equation With Logarithms

Example:

5ˣ = 17.

17 is not a convenient power of 5.

Take logs:

log(5ˣ) = log 17.

Power law:

x log 5 = log 17.

Therefore:

x = log 17 / log 5.

Keep the exact logarithmic form until a decimal is requested.

Solving Logarithmic Equations

Example:

log₂(x − 1) = 3.

Convert to exponential form:

x − 1 = 2³ = 8.

So:

x = 9.

Domain check:

x − 1 > 0 → x > 1.

x = 9 is valid.

Log Equations Can Produce Invalid Candidates

Consider:

log(x − 2) + log(x − 4) = log 5.

First state the domain:

x − 2 > 0 and x − 4 > 0 → x > 4.

Combine logs:

log[(x − 2)(x − 4)] = log 5.

Therefore:

(x − 2)(x − 4) = 5.

Expand:

x² − 6x + 8 = 5
x² − 6x + 3 = 0.

Solutions:

x = 3 ± √6.

Only values greater than 4 are allowed. Since 3 − √6 < 4, reject it. The valid answer is:

x = 3 + √6.

This connects logarithms back to surds and quadratic equations.

Why “Take Log of Both Sides” Is Legal

Logarithmic functions with a valid base are one-to-one on positive inputs.

So if A and B are positive and A = B, then:

log A = log B.

The operation preserves equality.

But the positivity requirement matters. You cannot casually apply real logarithms to expressions that may be zero or negative without checking the domain.

Exponential Growth Is Multiplicative, Not Additive

Linear growth adds roughly the same amount in equal time intervals.

Exponential growth multiplies by roughly the same factor in equal time intervals.

Model	Pattern
Linear	+5, +5, +5, +5
Exponential	×1.05, ×1.05, ×1.05, ×1.05

This difference is why compound interest, idealised population growth, radioactive decay and some cooling or depreciation models naturally use exponential functions.

A Standard Exponential Model

Q = Q₀aᵗ

where:

• Q₀ is the initial quantity;
• a is the multiplicative factor per time unit;
• t is time.

If a > 1, the model grows. If 0 < a < 1, the model decays.

For continuous-style models, e is often used:

Q = Q₀e^kt.

Here k > 0 represents growth and k < 0 represents decay.

Worked Example 2 — Find a Time

A quantity follows:

N = 120(1.08)ᵗ.

Find the time when N = 200.

Set:

200 = 120(1.08)ᵗ.

Divide:

5/3 = (1.08)ᵗ.

Take logs:

log(5/3) = t log(1.08).

Therefore:

t = log(5/3) / log(1.08).

Only now is a calculator needed for the decimal approximation.

A Model Is Not the World

Exponential models can become unrealistic if extended too far.

A population cannot grow exponentially forever in a finite environment. An investment return is not guaranteed to remain fixed. A biological decay model may contain multiple processes rather than one constant rate.

Use the model inside the conditions for which the model is justified.

Transforming an Exponential Relationship to a Straight Line

This idea connects forward to coordinate geometry and data modelling.

If:

y = kbˣ,

take logarithms:

log y = log k + x log b.

This has the straight-line form:

Y = c + mX

if we plot log y against x.

Then:

• gradient = log b;
• vertical intercept = log k.

This shows another purpose of logarithms: they can turn multiplicative exponential structure into additive linear structure.

Why Logarithms Compress Scale

Because logarithms convert multiplication into addition, huge multiplicative ranges become much smaller additive ranges.

For example:

10¹, 10², 10³, 10⁴

become:

1, 2, 3, 4

under base-10 logarithm.

This is why logarithmic ideas appear in scientific scales and data analysis, though the exact construction of each real-world scale must be learned on its own terms.

The Earliest Weak Link

What you see	Likely weak link
Cannot convert log₂8 = 3 to exponential form	definition of logarithm not secure
Writes log(a+b)=log a+log b	laws memorised without exponent-law origin
Solves log(x−4) equation and accepts x≤4	logarithm domain ignored
Uses decimal logs immediately in 3²ˣ=27	does not look for common-base structure first
Cannot explain asymptote x=0 for log graph	inverse-function relationship weak
Uses linear rather than exponential model for fixed percentage change	additive vs multiplicative change not distinguished

Common Mistakes to Repair

• log(a + b) = log a + log b. False in general.
• log(a/b) = log a / log b. False; quotient becomes subtraction.
• log(aⁿ) = (log a)ⁿ. False; exponent comes down as a multiplier.
• Dropping brackets around log arguments. Domain and algebra both depend on the whole argument.
• Forgetting x > 0 for log x. Real logs need positive arguments.
• Confusing log_ax with aˣ. They are inverse functions, not the same function.
• Rounding too early. Keep exact log ratios until the final numerical stage.
• Ignoring model units and domain. An equation without interpretation is not a complete modelling answer.

Retrieval Check

1. Rewrite 4³ = 64 in logarithmic form.
2. Rewrite log₅125 = 3 in exponential form.
3. State the domain and range of y = log₂x.
4. Simplify log₃9 + log₃3.
5. Express log₇20 using natural logarithms.
6. Solve 2ˣ = 32.
7. Explain why log(−4) has no real value.

Transfer Set

1. Solve 4ˣ⁺¹ = 8²ˣ⁻¹.
2. Solve 3ˣ = 10, giving an exact logarithmic expression before a decimal approximation.
3. Solve ln(x − 1) = 2.
4. Solve log₂x + log₂(x − 2) = 3, checking the domain.
5. Given y = 6(1.12)ᵗ, find the time when y = 15.
6. If y = kbˣ, show how a graph of log y against x can be used to determine k and b.

Answer outline — open only after attempting

1. 2²ˣ⁺² = 2⁶ˣ⁻³ → 2x+2=6x−3 → x=5/4.
2. x = log 10/log 3 = ln 10/ln 3.
3. x−1=e² → x=1+e².
4. Domain x>2. log₂[x(x−2)] = 3 → x²−2x=8 → x=4 or −2; valid x=4.
5. 15/6=(1.12)ᵗ → t=log(2.5)/log(1.12).
6. log y = log k + x log b; gradient m=log b so b=10ᵐ if base-10 logs are used; intercept c=log k so k=10ᶜ.

Independent Checks

• Convert your logarithmic answer back into exponential form.
• Check every logarithm argument is positive.
• For exponential equations, substitute the result into the original equation.
• For a growth model, check whether the result is larger than the initial quantity when the factor exceeds 1.
• For a decay model, check whether the quantity moves in the expected direction.
• Inspect calculator mode and parentheses before blaming the algebra.

For Parents and Tutors — What This Topic Is Really Testing

The strongest predictor of success here is not how many log laws the student can recite. It is whether the inverse relationship is stable.

Ask the student questions such as:

• “What question is log₃81 asking?”
• “Why does a product inside a log become addition?”
• “Why must the argument be positive?”
• “Why are the exponential and logarithmic graphs reflections?”
• “Why does fixed percentage growth lead to multiplication rather than addition?”

If those answers are clear, the formulas are usually recoverable. If those answers are missing, more repetitive simplification questions may only make the student faster at a fragile procedure.

The teaching target is reversible thinking: power → output, then output → required power.

Where This Connects Next

• Secondary 3 Additional Mathematics Topics Explained: The Complete A-Math Map
• Additional Mathematics Directory
• Secondary Math Tuition | Sec 3 Additional Mathematics Tutor
• Additional Mathematics | When the Method Works Until the Numbers Change

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Bukit Timah Tutor Mathematics
Do not begin with the log laws. Begin with the question the logarithm is answering. Then the laws become consequences rather than memory tests.