Additional Mathematics Classroom | Chapter 5: Exponential and Logarithmic Functions, Laws, Equations, Graphs and Models | SEC G3 K341

The 2027 SEC position

The official G3 K341 syllabus lists this topic as A6 Exponential and logarithmic functions. The required content includes exponential and logarithmic functions a^x, e^x, log_ax and ln x and their graphs; the laws of logarithms; the equivalence of y = a^x and x = log_ay; change of base; simplifying expressions and solving simple equations involving exponential and logarithmic functions; and using exponential and logarithmic functions as models.

This syllabus wording matters because the chapter is not only about “doing logs”. It has four connected jobs:

1. understand exponential and logarithmic functions as inverse relationships;
2. manipulate them accurately using index and logarithm laws;
3. solve equations by choosing a useful representation;
4. interpret exponential and logarithmic behaviour as a model of change.

Official reference: SEAB · 2027 SEC G3 Additional Mathematics K341.

The chapter is strongest when exponentials and logarithms are taught as one reversible system rather than two sets of formulas.

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What this chapter is really teaching

The visible topic is logarithms. The deeper capability is changing representation to expose a hidden operation.

An equation such as 2^x = 32 is easy because both sides can be written in the same base. But 2^x = 7 cannot be solved by elementary same-base recognition. A logarithm converts the exponent into an accessible algebraic quantity. Similarly, a product such as xy inside a logarithm becomes a sum of logarithms; a quotient becomes a difference; a power becomes a multiplier. Multiplicative structure becomes additive structure.

That change is the core intellectual move of the chapter. It is why logarithms historically made difficult numerical multiplication and division manageable, and it is why modern mathematics still uses them to reason about exponential processes, scale and rates.

• Exponential form asks: what value is produced after repeated multiplicative change?
• Logarithmic form asks: what exponent produced this value?
• Log laws ask: how can multiplicative relationships be rewritten additively?
• Graphs ask: what does the inverse relationship look like geometrically?
• Models ask: what does exponential growth or decay mean in a real context?

Chapter map

1. Index laws as prerequisite infrastructure
2. Exponential functions and repeated multiplication
3. The number e and natural exponential growth
4. Logarithms as inverse exponents
5. Domain, base conditions and notation
6. Laws of logarithms
7. Change of base
8. Solving exponential equations
9. Solving logarithmic equations
10. Substitution and disguised quadratics
11. Graphs and inverse symmetry
12. Exponential models
13. Logarithmic models and scale
14. Verification, restrictions and error diagnosis
15. SEC transfer questions
16. Original guided practice and worked answers

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1 · Index laws are the floor beneath the chapter

The old textbook begins with indices for good reason. Exponential functions are built from powers, and logarithm laws are mirrors of index laws. If index manipulation is unstable, logarithm work becomes a memory exercise with no structural anchor.

For a suitable non-zero base a, the familiar laws include:

• a^maⁿ = a^m+n;
• a^m/aⁿ = a^m−n;
• (a^m)ⁿ = a^mn;
• a⁰ = 1;
• a⁻ⁿ = 1/aⁿ;
• a^1/n = the nth root of a where the real-valued expression is defined.

These laws are not arbitrary rules. They come from counting repeated factors. For positive integers m and n, a^maⁿ contains m copies of a followed by n more copies, giving m + n copies in total. The more general laws extend that same structure consistently to zero, negative and fractional exponents.

Worked prerequisite A

Simplify 8^x · 4^1−x using base 2.

8^x = (2³)^x = 2^3x, and 4^1−x = (2²)^1−x = 2^2−2x.

Therefore the product is 2^3x+2−2x = 2^x+2.

Worked prerequisite B

Solve 9^x = 27.

Write both sides in base 3:

3^2x = 3³, so 2x = 3 and x = 3/2.

No logarithm is needed because a common base already exposes the exponent. Good method selection means using the simplest valid representation, not forcing every exponential equation through the log button.

The first diagnostic question

Before teaching logarithms, ask whether the learner can simplify a^2xa^3−x, rewrite 16^x in base 2, and interpret a negative exponent. If not, repair indices first. Logarithms should not be used to hide a weak exponent system.

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2 · Exponential functions: repeated proportional change

An exponential function has the variable in the exponent. A typical form is

y = a^x, where a > 0 and a ≠ 1.

Why the base conditions? If a = 1, the function is constantly 1 and has no exponential growth or decay behaviour. A positive base ensures the standard real-valued function is defined for every real x and stays positive. The restriction a ≠ 1 is also essential for an inverse logarithmic function.

If a > 1, the function grows as x increases. If 0 < a < 1, it decays. In both cases y remains positive. The graph passes through (0,1) because a⁰ = 1, and y = 0 is a horizontal asymptote because the function approaches zero but does not reach it.

Growth is multiplicative, not additive

Compare y = 3x + 2 with y = 2·3^x. In the linear function, equal increases in x add equal amounts to y. In the exponential function, equal increases in x multiply y by a constant factor. That difference explains why exponential growth can begin slowly and later become enormous.

If y = 5·2^x, increasing x by 1 doubles the output:

y(x + 1) = 5·2^x+1 = 2[5·2^x] = 2y(x).

The multiplier 2 is built into the base.

Decay

For y = 100(0.8)^x, each increase of 1 in x multiplies the value by 0.8. That means 80% remains after each step, equivalent to a 20% decrease per step. Exponential decay is still repeated multiplication; the multiplier is simply between 0 and 1.

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3 · The number e: the natural base

The SEC syllabus includes e^x and ln x. The constant e ≈ 2.71828… is the base of the natural exponential function. It appears naturally in continuous growth and calculus because e^x has the remarkable property that its derivative is itself.

At this stage, the important idea is not to memorise a long decimal for e. Treat e as an exact mathematical constant, just as π is exact. The notation ln x means logarithm to base e:

ln x = log_ex.

Therefore y = e^x and x = ln y are inverse statements.

Why ln appears later in calculus

Natural logarithms are not an arbitrary extra notation. They are the inverse language of e^x, and because e^x behaves so cleanly under differentiation, ln x becomes equally important when solving exponential equations and integrating certain forms. Learning the inverse relationship now prepares the calculus later.

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4 · Logarithms answer an exponent question

The statement

y = a^x

is equivalent to

x = log_ay,

where a > 0, a ≠ 1 and y > 0.

A logarithm therefore answers the question: what exponent must be placed on base a to produce y?

Examples as inverse translations

• 2⁵ = 32 means log₂32 = 5.
• 10⁻³ = 0.001 means log₁₀0.001 = −3.
• 9^1/2 = 3 means log₉3 = 1/2.
• e² = e² means ln(e²) = 2.

The fastest way to make logarithm notation natural is to translate repeatedly between the two forms until neither feels like the “new” form.

Special values

Because a⁰ = 1, log_a1 = 0.

Because a¹ = a, log_aa = 1.

These are not isolated properties to memorise. They are direct translations of index facts.

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5 · Domain and base restrictions are part of the mathematics

For the standard real logarithm log_ax, we require:

• a > 0;
• a ≠ 1;
• x > 0.

The last condition is especially important in equations. log(x − 3) is defined only when x − 3 > 0, so x > 3. An algebraic candidate that violates that condition is not a logarithmic solution.

Students sometimes treat restrictions as a final checklist. A stronger approach is to read them at the beginning. The domain can eliminate impossible branches before the algebra grows.

Worked domain check

For log₂(x − 1) + log₂(5 − x), both arguments must be positive:

x − 1 > 0 and 5 − x > 0.

Therefore 1 < x < 5.

Any later solution must lie inside this interval.

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6 · The laws of logarithms mirror the laws of indices

The three central log laws are:

• Product law: log_a(xy) = log_ax + log_ay.
• Quotient law: log_a(x/y) = log_ax − log_ay.
• Power law: log_a(x^r) = r log_ax.

These laws are valid where the logarithms involved are defined. Their structure comes directly from index laws.

Why the product law works

Let log_ax = m and log_ay = n. Then x = a^m and y = aⁿ. Multiplying:

xy = a^maⁿ = a^m+n.

Therefore log_a(xy) = m + n = log_ax + log_ay.

The log law is simply the index product law viewed through the inverse function.

Why “log of a sum” does not split

There is no general law saying log(x + y) = log x + log y. The right side corresponds to log(xy), not log(x + y). This is one of the most common serious misconceptions in the topic.

A numerical counterexample makes the error obvious. In base 10:

log 10 + log 10 = 1 + 1 = 2, while log(10 + 10) = log 20 ≠ 2.

The laws follow operations, not visual similarity.

Power-law misconception

log(x³) = 3 log x. It is not (log x)³. The exponent moves in front as a multiplier because the power law converts repeated multiplication inside the logarithm into repeated addition outside.

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7 · Simplifying logarithmic expressions

The most reliable sequence is: identify the common base, apply one law at a time, preserve the domain, and combine only where the structure allows.

Worked route A

Simplify 2 log₃x + log₃y − log₃z.

Use the power law:

2 log₃x = log₃x².

Then combine product and quotient:

log₃(x²y/z).

This assumes the original logarithms are defined, so x, y and z are positive.

Worked route B

Express 4 + log₃5 as a single logarithm.

Because log₃3 = 1, we can write

4 = 4 log₃3 = log₃3⁴ = log₃81.

Therefore

4 + log₃5 = log₃81 + log₃5 = log₃405.

Worked route C

Evaluate log₂8 + log₂4 − log₂16.

Directly: 3 + 2 − 4 = 1.

Or combine first:

log₂(8·4/16) = log₂2 = 1.

Multiple routes provide a useful verification opportunity.

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8 · Change of base: translating a logarithm into a calculator language

For valid bases a and c,

log_ab = log_cb / log_ca.

Choosing c = 10 gives log_ab = log b / log a. Choosing c = e gives log_ab = ln b / ln a.

The formula exists because calculators commonly provide base-10 log and natural log directly, while the mathematical problem may use another base.

Deriving change of base

Let x = log_ab. Then a^x = b. Take logarithms to any convenient base c:

log_c(a^x) = log_cb.

By the power law:

x log_ca = log_cb.

Hence x = log_cb/log_ca.

The formula is therefore another consequence of the inverse relationship and the power law.

Worked route

Evaluate log₅17 to three decimal places.

log₅17 = ln17/ln5 ≈ 1.760.

Check the scale: 5¹ = 5 and 5² = 25, so the exponent should lie between 1 and 2. The calculator value does.

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9 · Solving exponential equations: choose the route before calculating

There are several common routes. The strongest student identifies the structure first.

1. Same-base route: rewrite both sides with the same base and equate exponents.
2. Substitution route: if powers such as a^2x and a^x appear together, set u = a^x.
3. Logarithm route: when the bases cannot be made conveniently identical, take logarithms.

Worked route A · Same base

Solve 4^x+1 = 8^2x−1.

Write both sides in base 2:

2^2x+2 = 2^6x−3.

Therefore 2x + 2 = 6x − 3, so 5 = 4x and x = 5/4.

Worked route B · Logs

Solve 3^x = 10.

Take natural logs:

x ln3 = ln10.

Hence x = ln10/ln3 ≈ 2.096.

The exact logarithmic form is often more informative than an early decimal approximation.

Worked route C · A coefficient outside the exponential

Solve 5·2^x = 37.

First isolate the exponential:

2^x = 37/5.

Take logs:

x = ln(37/5)/ln2.

The order matters. Taking a logarithm before isolating the exponential is possible in some forms, but it often creates unnecessary clutter.

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10 · Disguised quadratics in exponential form

Some exponential equations become ordinary quadratics after substitution. This is one of the clearest examples of Additional Mathematics as a connected system: a new-looking problem is solved by recognising an old algebraic object.

Worked route A

Solve 9^x − 10·3^x + 9 = 0.

Since 9^x = (3^x)², let u = 3^x. Because 3^x > 0, u > 0.

The equation becomes

u² − 10u + 9 = 0

(u − 1)(u − 9) = 0.

So u = 1 or u = 9.

3^x = 1 gives x = 0; 3^x = 9 gives x = 2. Therefore x = 0 or x = 2.

Worked route B · Reject an impossible substitution value

Solve 4^x + 2^x − 6 = 0.

Let u = 2^x > 0. Then 4^x = u², so

u² + u − 6 = 0

(u + 3)(u − 2) = 0.

The quadratic gives u = −3 or 2, but u = 2^x cannot be negative. Hence u = 2 and x = 1.

The substitution carries a domain. Writing u = 2^x without recording u > 0 invites an impossible branch.

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11 · Solving logarithmic equations

There are two main strategies:

• combine logarithms until a single logarithm remains, then convert to exponential form;
• if logarithms with the same base are equal, equate their positive arguments.

Domain restrictions must accompany either route.

Worked route A · Equality of logarithms

Solve log₃(2x − 1) = log₃7.

The logarithm function is one-to-one, so equal logs with the same valid base have equal positive arguments:

2x − 1 = 7.

Hence x = 4. The argument 2x − 1 = 7 is positive, so the solution is valid.

Worked route B · Combine first

Solve log₂x + log₂(x − 2) = 3.

Domain: x > 2.

Combine:

log₂[x(x − 2)] = 3.

Convert to exponential form:

x(x − 2) = 2³ = 8.

x² − 2x − 8 = 0

(x − 4)(x + 2) = 0.

The algebra gives x = 4 or −2, but the domain requires x > 2. Therefore x = 4.

Worked route C · A constant and a logarithm

Solve 2 + log₅x = log₅(3x + 20).

Write 2 = log₅25. Then

log₅25 + log₅x = log₅(3x + 20).

So log₅(25x) = log₅(3x + 20).

Hence 25x = 3x + 20, giving 22x = 20 and x = 10/11.

Both log arguments are positive at x = 10/11, so the solution is valid.

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12 · Logs on both sides do not remove domain responsibility

Suppose an equation simplifies to log(x − 1) = log(5 − x). It is tempting to equate x − 1 = 5 − x immediately. That algebra is fine only if both arguments are in the logarithm’s domain.

We require x − 1 > 0 and 5 − x > 0, so 1 < x < 5. Equating arguments gives 2x = 6 and x = 3, which lies in the domain. Therefore x = 3 is valid.

In harder equations, a quadratic may produce several candidates. Domain checks are what distinguish algebraic candidates from logarithmic solutions.

A common hidden failure

When students combine log(x − 2) + log(x + 1), they may write log(x² − x − 2) and then assume only the product needs to be positive. That loses information. The original expression requires both x − 2 > 0 and x + 1 > 0, which together give x > 2. A positive product alone would also allow x < −1, but the individual logarithms would not exist there.

Algebraic simplification does not erase the domain conditions of the original expression.

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13 · Graphs: exponentials and logarithms are mirror images

If y = a^x and x = log_ay are equivalent, then the logarithmic function y = log_ax is the inverse of the exponential function y = a^x. Inverse functions have graphs reflected in the line y = x.

For a > 1:

• y = a^x has domain all real x and range y > 0;
• y = log_ax has domain x > 0 and range all real y;
• (0,1) on the exponential corresponds to (1,0) on the logarithm;
• the exponential has horizontal asymptote y = 0;
• the logarithm has vertical asymptote x = 0.

The domain and range swap because the input and output swap under inversion. The asymptotes also exchange orientation under reflection in y = x.

Why the graph never crosses the asymptote

a^x is always positive for positive base a. It may become arbitrarily close to zero for suitable x, but it cannot equal zero. Therefore y = 0 is approached but not reached. The inverse logarithmic graph similarly approaches x = 0 from the positive side but cannot cross into x ≤ 0 because logarithms of non-positive real numbers are outside the standard real domain.

Transformations

Although the syllabus focuses on the functions and their graphs rather than an enormous catalogue of graph transformations, students should still interpret basic changes. For example, y = 2^x + 3 is y = 2^x shifted upward by 3, so its horizontal asymptote becomes y = 3. Similarly, y = ln(x − 4) shifts y = ln x right by 4, moving the vertical asymptote to x = 4.

Graph transformations should be read from the function structure, not memorised as disconnected sketches.

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14 · Exponential models: repeated percentage change

The SEC syllabus explicitly includes using exponential and logarithmic functions as models. A common model is

Q = Q₀a^t

or, for continuous growth or decay,

Q = Q₀e^kt.

Q₀ is the initial quantity. The base or parameter controls the multiplicative change. The variable t often represents time.

Worked model A · Compound growth

A population is modelled by P = 1200(1.06)^t, where t is measured in years.

• Initial population: P(0) = 1200.
• Annual multiplication factor: 1.06.
• Modelled annual growth rate: 6%.

To find when the population first reaches 2000 in the continuous algebraic model, solve

1200(1.06)^t = 2000.

(1.06)^t = 5/3.

Take logs:

t = ln(5/3)/ln(1.06) ≈ 8.77 years.

If the context requires whole yearly observations, interpretation may require the ninth year rather than reporting 8.77 as though the population were counted continuously. The model and the reporting rule both matter.

Worked model B · Decay

A quantity follows M = 500e^−0.18t.

At t = 0, M = 500. Since the exponent coefficient is negative, the model decays. To find the time when M = 200:

500e^−0.18t = 200

e^−0.18t = 0.4

−0.18t = ln0.4

t = −ln0.4/0.18 ≈ 5.09.

The negative logarithm of 0.4 and the negative coefficient combine to produce a positive time, as expected.

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15 · Doubling time and half-life

Logarithms are especially useful when the unknown is time.

For Q = Q₀e^kt with k > 0, doubling time T satisfies

2Q₀ = Q₀e^kT.

So 2 = e^kT, hence

T = ln2/k.

For decay Q = Q₀e^−kt with k > 0, half-life H satisfies

1/2 = e^−kH, so

H = ln2/k.

The same ln2 appears because doubling and halving are reciprocal multiplicative thresholds.

Why the initial amount cancels

Doubling time depends on the growth rate, not the starting amount. Whether the initial quantity is 5, 500 or 5 million, the equation divides by Q₀ and the same time emerges. This is a characteristic feature of exponential proportional growth.

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16 · Logarithmic models: measuring multiplicative scale

Logarithms are useful when a quantity spans many orders of magnitude. A logarithmic scale compresses multiplicative differences into additive steps. This is why logarithms appear in scales used for sound intensity, acidity and earthquake magnitude, among many other contexts.

The old textbook uses earthquake magnitude as an application context. The modern teaching point remains valuable: a difference of one unit on a logarithmic scale does not necessarily mean “one more” in the original physical quantity. It may represent multiplication by a fixed factor.

If a model has the form

M = log₁₀(I/I₀),

then increasing M by 1 multiplies I/I₀ by 10. Increasing M by 2 multiplies it by 100. The additive scale is recording multiplicative change.

When teaching real-world logarithmic scales, use the equation actually supplied in the question rather than assuming every scale uses the same physical definition or factor. The mathematical interpretation follows the stated model.

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17 · Linearising an exponential model

A useful bridge to coordinate geometry and straight-line transformation is the model

y = kb^x.

Taking logarithms:

ln y = ln k + x ln b.

If we plot Y = ln y against X = x, the relationship is linear:

Y = (ln b)X + ln k.

The gradient gives ln b and the intercept gives ln k. Therefore b = e^gradient and k = e^intercept.

This is an important example of why logarithms are more than equation-solving tools. They can transform nonlinear multiplicative relationships into linear additive ones that are easier to analyse.

Worked route

Suppose ln y plotted against x gives a straight line with equation

ln y = 0.7x + 1.2.

Comparing with ln y = x ln b + ln k:

ln b = 0.7, so b = e^0.7.

ln k = 1.2, so k = e^1.2.

Hence the original model is y = e^1.2(e^0.7)^x, equivalently y = e^1.2+0.7x.

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18 · The error map: where logarithm work actually breaks

“Weak at logs” is too vague to guide teaching. The first weak link may be one of several different things.

• Index-law failure: exponents are simplified incorrectly before logarithms even appear.
• Inverse-relationship failure: the learner cannot move between y = a^x and x = log_ay.
• Domain failure: zero or negative log arguments are accepted.
• Law-selection failure: a product law is incorrectly used on a sum.
• Power-law failure: log(x^r) is confused with (log x)^r.
• Change-of-base failure: numerator and denominator are reversed.
• Equation-routing failure: a same-base equation is made unnecessarily complicated with logs, or a non-common-base equation is attacked by guesswork.
• Substitution failure: a^2x is not recognised as (a^x)².
• Candidate filtering failure: solutions of the algebraic equation are not checked against original log domains.
• Model interpretation failure: the learner calculates time or rate but does not explain its contextual meaning.

Different failures require different repairs. A student who knows the log laws but cannot solve 4^x − 5·2^x + 4 = 0 needs substitution and quadratic recognition, not another page of product-law exercises. A student who finds x = −2 from a log equation and accepts it despite log(x + 1) is showing domain failure, not algebra failure.

The first weak-link protocol

1. Can the learner manipulate indices accurately?
2. Can the learner translate exponential ↔ logarithmic form?
3. Can the learner state the log domain?
4. Can the learner choose the correct log law?
5. Can the learner identify the correct equation route?
6. Can the learner verify the result in the original expression or model?

Stop at the first unstable step. Repair that step, then return to the original problem. This gives evidence that the repair transferred.

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19 · SEC examination transfer

A strong G3 chapter should not end with isolated exercises labelled “logarithms”. The examination objective is broader: identify the relevant concept, connect forms, choose a method, interpret results and communicate mathematically.

A complete practice architecture therefore needs several layers:

1. Direct fluency: index laws, log laws, simple translations and same-base equations.
2. Representation choice: decide between same-base, logarithm and substitution routes.
3. Domain control: solve log equations with multiple candidates and restrictions.
4. Graph interpretation: connect inverse functions, asymptotes, intercepts and transformations.
5. Modelling: construct or use exponential models and interpret parameters.
6. Mixed-topic work: combine logs with quadratics, coordinate geometry or later calculus.

The progression should remove chapter labels gradually. The learner must eventually recognise an exponential object because of its structure, not because the worksheet says “Chapter 5”.

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20 · Original guided practice

The following questions are original to this classroom guide. They progress from prerequisite structure into equations, graphs, models and mixed reasoning. Complete the first pass without the worked answers. Mark the first line at which the method becomes uncertain.

A · Indices and exponential structure

1. Simplify 2^3x·4^1−x.
2. Solve 8^x = 4^x+1.
3. Solve 9^x−1 = 27.
4. Solve 5^2x·5^1−x = 125.
5. Explain why a^x is always positive for a > 0.

B · Logarithmic form and laws

6. Write 3⁴ = 81 in logarithmic form.
7. Write log₅(1/25) = −2 in exponential form.
8. Simplify log₂8 + log₂4 − log₂16.
9. Express 3 log₇x − 2 log₇y as a single logarithm.
10. Express 2 + log₃5 as a single logarithm.
11. Explain why log(x + y) cannot generally be split into log x + log y.

C · Change of base and evaluation

12. Express log₇20 using natural logarithms.
13. Evaluate log₄11 to three decimal places.
14. Show that log_ab = 1/log_ba.
15. If log₂3 = p, express log₃2 in terms of p.

D · Exponential equations

16. Solve 4^x+1 = 8^2x−1.
17. Solve 3^x = 14, giving an exact logarithmic answer and a decimal to three significant figures.
18. Solve 7·2^x = 50.
19. Solve 9^x − 10·3^x + 9 = 0.
20. Solve 4^x + 2^x − 6 = 0.
21. Solve 25^x − 6·5^x + 5 = 0.

E · Logarithmic equations

22. Solve log₃(2x − 1) = 2.
23. Solve log₂x + log₂(x − 2) = 3.
24. Solve log₅(x + 4) − log₅x = 1.
25. Solve 2 + log₅x = log₅(3x + 20).
26. Solve ln(x − 1) + ln(x + 1) = ln8.
27. Solve log₁₀(x − 2) = 1 − log₁₀x.

F · Graphs and models

28. State the domain, range and horizontal asymptote of y = 3^x.
29. State the domain, range and vertical asymptote of y = log₃x.
30. Explain why y = 3^x and y = log₃x are reflections in y = x.
31. A population follows P = 800(1.04)^t. Find the modelled population after 10 years.
32. For P = 800(1.04)^t, find when P reaches 1200.
33. A quantity follows Q = 600e^−0.12t. Find the time at which Q = 300.
34. For Q = Q₀e^kt, derive the doubling time when k > 0.
35. If ln y = 0.4x + 1.5, express y in the form kb^x.

G · Mixed reasoning and diagnosis

36. Solve 2^2x − 5·2^x + 4 = 0.
37. Solve log₂(x − 1) + log₂(5 − x) = 1.
38. Find k if 3^x = k has solution x = 2.5.
39. A student writes log(x + 4) = log x + log4. Explain the error using a numerical counterexample.
40. A student solves log(x − 2) + log(x + 1) = 1 and accepts a value x < −1 because the product (x − 2)(x + 1) is positive. Explain why this is invalid.
41. The graph of y = 2^x + c has horizontal asymptote y = 5. Find c.
42. The graph y = ln(x − a) passes through (a + 1,0). Explain why this is always true.
43. Find the exact x-coordinate of the intersection of y = e^x and y = 7.
44. If 5^x = 2, show that 25^x = 4 without finding x.
45. Explain when it is better to use a common-base method rather than logarithms.

21 · Worked answers and reasoning checkpoints

1. 2^3x·4^1−x = 2^3x·2^2−2x = 2^x+2.
2. 2^3x = 2^2x+2, so 3x = 2x + 2 and x = 2.
3. 3^2x−2 = 3³, so 2x − 2 = 3 and x = 5/2.
4. 5^2x+1−x = 5³, so x + 1 = 3 and x = 2.
5. For a > 0, every real power a^x is positive; the exponential graph approaches zero but does not cross it.
6. log₃81 = 4.
7. 5⁻² = 1/25.
8. 3 + 2 − 4 = 1.
9. 3 log₇x − 2 log₇y = log₇(x³/y²).
10. 2 = log₃9, so result = log₃9 + log₃5 = log₃45.
11. The sum law does not exist. For example, log₁₀10 + log₁₀10 = 2, but log₁₀20 ≠ 2.
12. ln20/ln7.
13. log₄11 = ln11/ln4 ≈ 1.730.
14. log_ab = ln b/ln a and log_ba = ln a/ln b, so they are reciprocals: log_ab = 1/log_ba.
15. 1/p.
16. x = 5/4.
17. x = ln14/ln3 ≈ 2.40.
18. 2^x = 50/7, so x = ln(50/7)/ln2.
19. Let u = 3^x > 0. u² − 10u + 9 = 0 gives u = 1 or 9. Hence x = 0 or 2.
20. Let u = 2^x > 0. u² + u − 6 = 0 gives u = 2 or −3. Hence x = 1.
21. Let u = 5^x > 0. u² − 6u + 5 = 0 gives u = 1 or 5. Hence x = 0 or 1.
22. 2x − 1 = 3² = 9, so x = 5.
23. Domain x > 2. x(x − 2) = 8 gives x = 4 or −2. Hence x = 4.
24. Domain x > 0. log₅[(x + 4)/x] = 1 gives (x + 4)/x = 5, so 4 = 4x and x = 1.
25. x = 10/11.
26. Domain x > 1. ln[(x − 1)(x + 1)] = ln8 gives x² − 1 = 8, so x² = 9. Only x = 3 satisfies the domain. x = 3.
27. Domain x > 2. Write 1 = log10. Then log[x(x − 2)] = 1 gives x² − 2x = 10, so x = 1 ± √11. Only x = 1 + √11 satisfies x > 2.
28. Domain all real x; range y > 0; horizontal asymptote y = 0.
29. Domain x > 0; range all real y; vertical asymptote x = 0.
30. They are inverse functions. Inverse graphs exchange x- and y-coordinates, which is reflection in y = x.
31. P = 800(1.04)¹⁰ ≈ 1184 to the nearest whole number.
32. (1.04)^t = 1.5, so t = ln1.5/ln1.04 ≈ 10.34 years.
33. 300 = 600e^−0.12t gives 1/2 = e^−0.12t. Hence t = ln2/0.12 ≈ 5.78.
34. 2Q₀ = Q₀e^kT gives 2 = e^kT, so T = ln2/k.
35. ln y = 0.4x + 1.5 gives y = e^1.5e^0.4x = e^1.5(e^0.4)^x.
36. Let u = 2^x > 0. u² − 5u + 4 = 0 gives u = 1 or 4. Therefore x = 0 or 2.
37. Domain requires x − 1 > 0 and 5 − x > 0, so 1 < x < 5. Combine: (x − 1)(5 − x) = 2. This gives −x² + 6x − 5 = 2, or x² − 6x + 7 = 0. Thus x = 3 ± √2, both within (1,5).
38. If x = 2.5, then k = 3^2.5 = 9√3.
39. Take x = 1: log(1 + 4) = log5, while log1 + log4 = log4. Since log5 ≠ log4, the proposed law is false.
40. The original logarithms require x − 2 > 0 and x + 1 > 0, so x > 2. A positive product for x < −1 does not make the individual log arguments positive.
41. The horizontal asymptote of y = 2^x + c is y = c. Therefore c = 5.
42. At x = a + 1, x − a = 1 and ln1 = 0. Hence the point (a + 1,0) always lies on the graph.
43. e^x = 7 gives x = ln7.
44. 25^x = (5²)^x = (5^x)² = 2² = 4.
45. Use a common base when both sides can be rewritten cleanly in the same base; it is usually shorter and keeps the solution exact without introducing logarithms unnecessarily.

Diagnostic note: Question 37 deliberately keeps both solutions 3 − √2 and 3 + √2 because both lie in the original domain 1 < x < 5. Domain checking should remove invalid roots, not automatically remove one root because a quadratic produced two.

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22 · A two-week revision architecture

Exponential and logarithmic work is particularly vulnerable to “method familiarity”. Immediately after a lesson, students know that the page is about logs. Two weeks later, an exponential equation inside a mixed paper may not trigger the same method. Revision must therefore train recognition as well as execution.

1. Day 1: index-law repair and same-base exponential equations.
2. Day 2: exponential ↔ logarithmic translation and domain conditions.
3. Day 3: product, quotient and power laws with misconceptions contrasted explicitly.
4. Day 5: change of base and non-common-base equations.
5. Day 7: logarithmic equations with domain filtering.
6. Day 9: substitution equations such as a^2x + ba^x + c = 0.
7. Day 11: graphs, inverse functions and asymptotes.
8. Day 14: models and a mixed unlabelled diagnostic set.

The mixed set is essential. Include some questions where logarithms are unnecessary, some where they are the shortest route, and some where a quadratic substitution is better. Method selection grows only when alternatives are present.

Five-minute retrieval test

• Write the equivalence y = a^x ↔ x = log_ay.
• State the three log laws.
• State the domain of log_ax.
• Write the change-of-base formula.
• Name three routes for solving exponential equations.
• Explain why log(x + y) does not split.
• State the asymptotes of y = a^x and y = log_ax.

23 · For teachers: make route choice part of the lesson

Many logarithm lessons become lists of procedures: law 1, law 2, law 3, change of base, solve. That can produce competent worksheet performance but weak transfer. A better routine is to show several equations together and ask students to classify them before solving.

• 9^x = 27 → same base.
• 3^x = 11 → logarithm.
• 9^x − 10·3^x + 9 = 0 → substitution.
• log x + log(x − 2) = log8 → combine logs, then solve and filter by domain.

The classification step is where method selection is learned. Once students can name the structure, the algebra often becomes ordinary.

Teach laws with counterexamples

Put valid and invalid statements side by side:

• log(xy) = log x + log y — valid.
• log(x + y) = log x + log y — false.
• log(x^r) = r log x — valid where defined.
• log(x^r) = (log x)^r — false.

Students remember boundaries more reliably when the neighbouring wrong move is made visible.

Connect the graph to the algebra

When introducing log domain x > 0, show it on the graph. When discussing a^x > 0, show the horizontal asymptote. When converting y = a^x into x = log_ay, reflect the graph. The learner should not have separate memories for formula, domain and picture. They describe the same function system.

24 · For parents: what real progress sounds like

A student who is relying on memory may say, “When I see log I use the log laws.” A student with stronger structural control begins to say:

• “Both sides can be written in base 3, so I do not need logs.”
• “The exponent is unknown and the bases do not match, so I will take logs.”
• “This is quadratic in 2^x, so I will substitute u = 2^x.”
• “The log argument must stay positive, so this root is impossible.”
• “The product law works because multiplication of powers adds exponents.”
• “The logarithm graph is the inverse of the exponential graph.”
• “The model grows by a constant factor, not a constant amount.”

This shift in language is evidence that the learner is carrying the structure of the subject. It is more valuable than being able to repeat a single model solution quickly.

25 · What carries forward

This chapter is foundational for later G3 work. Exponential and logarithmic functions reappear in differentiation, integration, modelling, straight-line transformation and mixed algebra. The natural exponential e^x becomes especially important in calculus because differentiation and integration interact with it in unusually clean ways. Logarithmic differentiation is beyond this immediate syllabus job, but the inverse relationship and law control established here support later mathematical study.

The old textbook’s next chapter is Coordinate Geometry in Two Dimensions. That remains strongly relevant to the 2027 G3 route, but the modern SEC crosswalk must inspect which circle, line, transformation and linear-law components belong in current K341 and which are assumed from G3 Mathematics. The next chapter will therefore be rebuilt from that legacy geometry chapter rather than copied page for page.

The conversion rule remains unchanged: preserve useful teaching, make the dependency visible, and let SEAB 2027 control the examinable boundary.

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Official syllabus reference

Singapore Examinations and Assessment Board · 2027 SEC G3 Additional Mathematics K341

Curriculum and assessment requirements can change. The official SEAB syllabus remains the controlling source for current subject codes, examinable content and examination structure.

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