Additional Mathematics Classroom | Chapter 7: Linear Law, Straight-Line Transformations, Parameter Recovery and Model Testing | SEC G3 K341

The 2027 SEC position

In the 2027 G3 Additional Mathematics syllabus K341, Linear Law sits inside G2 Coordinate geometry in two dimensions. The syllabus requires students to transform given relationships—including y = axⁿ and y = kb^x—to linear form in order to determine unknown constants from a straight-line graph.

This requirement does not appear in the 2027 G2 Additional Mathematics K232 coordinate-geometry content. The chapter is therefore part of the G3 K341 route.

The important wording is “transform”, “linear form”, “determine unknown constants” and “straight-line graph”. Those four phrases define the actual examination job. A student who can mechanically take logarithms but cannot identify the axes or recover the original constants has not completed the topic.

Linear Law is not about drawing a straight line. It is about constructing the representation in which the unknown constants become measurable.

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

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What this chapter is really teaching

Linear Law teaches representation engineering. The relationship is not changed into a different physical law. The same relationship is rewritten using transformed variables so that straight-line geometry can expose its parameters.

Suppose two variables obey a nonlinear relationship. The raw graph may curve. Estimating two unknown constants directly from a curve can be difficult. But if an algebraic transformation produces

Y = mX + c,

then a straight-line graph gives two robust quantities:

• gradient m;
• vertical intercept c.

If m and c have been defined in terms of the original unknown constants, the graph becomes a measuring instrument for those constants.

This requires the learner to move through five layers:

1. Original relationship: what nonlinear family are x and y believed to follow?
2. Transformation: what operation makes the relationship linear?
3. New variables: what are X and Y?
4. Graph evidence: what are m and c?
5. Return path: what do m and c imply about the original constants?

The fifth layer is where many students lose marks. They identify the transformed gradient correctly and then forget that the original parameter may be the antilogarithm of that gradient rather than the gradient itself.

Chapter map

1. Why nonlinear data are linearised
2. The target architecture Y = mX + c
3. Choosing X and Y
4. Power law y = axⁿ
5. Exponential law y = kb^x
6. Natural exponential law y = ae^bx
7. Reverse recovery from a transformed graph
8. Transformations without logarithms
9. Multiple valid linear forms
10. Tables of transformed data
11. Gradient, intercept and parameter recovery
12. Line of best fit and anomalous data
13. Model testing and residual thinking
14. Interpolation and cautious extrapolation
15. Error diagnosis
16. SEC transfer
17. Original guided practice
18. Worked answers and mastery checkpoint

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1 · Why study Linear Law?

Imagine a physics experiment in which a quantity y is measured for several values of x. The points form a curve. The experimenter suspects that the variables follow some model containing two unknown constants. A curve can display the trend, but reading two constants accurately from it is not always easy.

A straight line is different. Its geometry is controlled by two quantities: gradient and intercept. If the nonlinear model can be transformed into straight-line form, those two graph features can encode the two unknown constants.

Historically, this made linearisation a powerful scientific method. Even with modern nonlinear regression software, the educational value remains substantial because linearisation makes model structure visible. It forces us to understand how the parameters affect the relationship rather than asking software to fit a curve without explanation.

A pendulum-style idea

Suppose an experiment suggests that period T and length L obey a power law

T = kLⁿ.

The raw T-against-L graph is curved when n ≠ 1. Taking logarithms gives

log T = log k + n log L.

Now a graph of log T against log L is straight:

• gradient = n;
• intercept = log k.

The exponent n is visible directly from the slope. The scale factor k is recovered from the intercept by reversing the logarithm.

The straight-line graph has converted a nonlinear model into parameter evidence.

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2 · The target is always a straight-line equation

The standard target is

Y = mX + c.

The capital letters matter. X and Y may be transformed versions of the original x and y. For example:

• X = x, Y = ln y;
• X = log x, Y = log y;
• X = 1/x, Y = y;
• X = x², Y = xy;
• X = x, Y = y/x.

There is nothing magical about the letters. They simply help us separate the variables plotted on the straight-line graph from the original variables.

The complete matching statement

A good Linear Law solution should make four identifications explicit:

1. Y = …
2. X = …
3. m = …
4. c = …

For example, from

ln y = bx + ln a,

we identify:

• Y = ln y;
• X = x;
• m = b;
• c = ln a.

Therefore the required plot is ln y against x, not y against x and not x against ln y.

“Y against X” has an order

In graph language, “plot Y against X” means:

• vertical axis: Y;
• horizontal axis: X.

Reversing the axes changes the gradient and intercept. A correct transformation paired with reversed axes is not a complete correct solution.

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3 · Power law: y = axⁿ

This is one of the explicit SEC forms. Let

y = axⁿ,

where a and n are constants and the logarithms involved are defined.

Take logarithms:

log y = log(axⁿ)

= log a + log(xⁿ)

= log a + n log x.

Rearrange to match straight-line order:

log y = n log x + log a.

Therefore:

• Y = log y;
• X = log x;
• gradient m = n;
• intercept c = log a.

After obtaining c from the graph, recover a using the same logarithm base. If common logarithms are used, a = 10^c. If natural logarithms are used, a = e^c.

Worked Example 1 · Recover a power-law model

A straight-line graph of log₁₀ y against log₁₀ x has gradient 1.6 and vertical intercept 0.3010. Given that y = axⁿ, find a and n.

From the transformed equation

log y = n log x + log a,

we have n = 1.6 and log a = 0.3010.

Therefore a = 10^0.3010 ≈ 2.

The recovered model is approximately

y = 2x^1.6.

The gradient gave n directly. The intercept did not give a directly; it gave log a.

Worked Example 2 · Build the plot from the model

The variables x and y are believed to satisfy y = 5x^2.4. What straight-line graph would represent this model?

Take logs:

log y = 2.4 log x + log5.

Plot log y against log x.

• gradient = 2.4;
• intercept = log5.

This forward direction matters because students must be able to design the transformation, not only read one already supplied.

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4 · Exponential law: y = kb^x

The second explicit SEC family is

y = kb^x,

where k and b are constants with the usual positive-base conditions required for real logarithms.

Take logarithms:

log y = log(kb^x)

= log k + x log b.

So

log y = (log b)x + log k.

Therefore:

• Y = log y;
• X = x;
• gradient m = log b;
• intercept c = log k.

Both parameters now require a return step:

• b = antilog(m);
• k = antilog(c).

Worked Example 3 · Gradient is not b

A graph of log₁₀ y against x for y = kb^x has gradient 0.4771 and vertical intercept 1.0000. Find k and b.

log b = 0.4771, so b ≈ 10^0.4771 ≈ 3.

log k = 1, so k = 10.

Hence

y = 10·3^x.

The classic mistake is b = 0.4771. That confuses the transformed gradient with the original exponential base.

Growth and decay appear in the gradient

If b > 1, then log b > 0 and the transformed graph has positive gradient. If 0 < b < 1, then log b < 0 and the transformed graph has negative gradient. The sign of the straight-line gradient therefore records whether the original exponential model grows or decays.

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5 · Natural exponential law: y = ae^bx

The old textbook gives this form particular attention because natural logarithms make it especially clean:

y = ae^bx.

Take ln:

ln y = ln a + ln(e^bx)

= ln a + bx.

So

ln y = bx + ln a.

Thus a graph of ln y against x has:

• gradient = b;
• intercept = ln a.

The exponent coefficient b is visible directly. The initial scale a must be recovered as e^intercept.

Worked Example 4 · Exponential decay

A graph of ln y against x is a straight line through (0, ln12) with gradient −0.35. Given y = ae^bx, find the model.

b = −0.35.

ln a = ln12, so a = 12.

Therefore

y = 12e^−0.35x.

The negative gradient of the transformed graph corresponds to decay in the original model.

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6 · Reverse direction: from a straight-line graph back to the nonlinear model

Many examination questions begin with the graph rather than the original transformation. The learner may be told that plotting one transformed quantity against another produces a straight line through two given points. The job is to use those graph points to recover gradient and intercept, then map them back into the original model.

Worked Example 5 · Two transformed graph points

The relationship is y = ae^bx. A graph of ln y against x passes through (1,4) and (5,12). Find a and b.

For Y = ln y and X = x:

gradient b = (12 − 4)/(5 − 1) = 8/4 = 2.

So the straight-line equation is

ln y = 2x + c.

Using (1,4):

4 = 2 + c, so c = 2.

But c = ln a. Therefore a = e².

The model is

y = e²e^2x = e^2x+2.

Notice the layered interpretation: graph intercept c = 2, but original parameter a = e².

Worked Example 6 · Power law from two transformed points

For y = axⁿ, the graph of log y against log x passes through (0,0.6021) and (2,3.6021), using base-10 logarithms. Find a and n.

Gradient n = (3.6021 − 0.6021)/(2 − 0) = 3/2 = 1.5.

Intercept log a = 0.6021.

a ≈ 10^0.6021 ≈ 4.

Thus y = 4x^1.5, equivalently y = 4x^3/2.

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7 · Transformations without logarithms

The SEC wording says “including” y = axⁿ and y = kb^x. The old textbook usefully demonstrates that Linear Law is a broader representation idea: some nonlinear-looking equations become linear after algebraic rearrangement, reciprocals, multiplication or division rather than logarithms.

These examples strengthen the central skill but should support, not obscure, the explicit syllabus forms.

Example A · y = a + b/x

Write

y = b(1/x) + a.

Let Y = y and X = 1/x. Then

Y = bX + a.

Plot y against 1/x:

• gradient = b;
• intercept = a.

Example B · y = ax² + bx

For x ≠ 0, divide by x:

y/x = ax + b.

Let Y = y/x and X = x. Then

Y = aX + b.

Plot y/x against x. Gradient = a, intercept = b.

Example C · y = x³ + ax + b

Subtract the known nonlinear part:

y − x³ = ax + b.

Let Y = y − x³ and X = x. Then

Y = aX + b.

The transformation is not “take logs because this is Linear Law”. It is “rearrange until the unknown constants occupy gradient and intercept positions”.

Example D · xy = ax² + b

Let Y = xy and X = x². Then

Y = aX + b.

Plot xy against x². The algebra itself suggests the transformed axes.

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8 · There can be more than one valid linearisation

An important idea in the old textbook is that a nonlinear equation may sometimes be rearranged into more than one straight-line form. This is mathematically useful because it teaches that transformations are choices, not rituals.

Suppose an equation can be manipulated into either

Y₁ = m₁X₁ + c₁

or

Y₂ = m₂X₂ + c₂.

Both may be correct. The better choice depends on the available data, ease of calculating transformed values, clarity of parameter recovery and numerical stability.

The decision question

Do not ask only, “Can this be made linear?” Ask, “Which valid linear form makes the graph and parameter interpretation simplest?”

This is a 21st-century mathematical habit. Software gives us many ways to fit or transform data. Good modelling still requires choosing a representation that is interpretable and appropriate.

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9 · Transforming a data table

In a full Linear Law question, the transformation often has to be applied to experimental data before the graph can be drawn.

Suppose x and y are believed to satisfy

y = ax² + bx.

For x ≠ 0:

y/x = ax + b.

The required transformed table needs columns for:

• x;
• y;
• Y = y/x.

The plotted graph is Y against X = x.

Worked Example 7 · Transform a small table

Suppose the observed values are:

x	2	4	6	8
y	12	40	84	144

Calculate y/x:

x	2	4	6	8
y/x	6	10	14	18

The transformed points lie on Y = 2X + 2. Therefore a = 2 and b = 2. The original model is

y = 2x² + 2x.

Check x = 6: 2(36) + 12 = 84, matching the table.

Rounding transformed values

When transformed values require logarithms, reciprocals or ratios, keep enough significant figures so the plotted graph is not distorted by premature rounding. The old textbook’s graphing practice is still useful here: transformed data are measurements, and careless rounding changes the evidence the graph is supposed to reveal.

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10 · Gradient should be measured from the line, not merely from adjacent data points

If experimental points do not lie perfectly on one line, a line of best fit is drawn. The gradient should then be estimated using two well-separated points on the fitted line, preferably not simply two original data points unless they lie on that line.

Why use well-separated points? Because a small reading error in coordinates creates a smaller proportional error in the gradient when the horizontal and vertical differences are large.

Worked Example 8 · Recover parameters from a fitted line

A graph of ln y against x is fitted by a straight line. Two convenient points on the fitted line are (1.0,2.2) and (5.0,4.6).

Gradient m = (4.6 − 2.2)/(5.0 − 1.0) = 2.4/4 = 0.6.

Using Y = mX + c and (1,2.2):

2.2 = 0.6(1) + c, so c = 1.6.

If the original model is y = ae^bx:

• b ≈ 0.6;
• ln a ≈ 1.6, so a ≈ e^1.6 ≈ 4.95.

The approximate recovered model is

y ≈ 4.95e^0.6x.

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11 · Anomalous points: a straight-line graph is also a diagnostic

The old textbook includes experimental sets in which one value may be inaccurate. This is pedagogically valuable because Linear Law is not only a parameter-recovery technique; it can also expose data that do not behave like the rest of the observations.

If most transformed points lie close to a straight line and one lies far away, several explanations are possible:

• measurement error;
• recording or transcription error;
• calculation error in the transformed value;
• the experimental condition changed;
• the proposed model is only approximately valid and the point lies in a region where it breaks down;
• ordinary random variation.

Do not automatically delete a point because it is inconvenient. A graph can flag an anomaly; mathematical and experimental reasoning must decide what the anomaly means.

Estimating a corrected value

If a question explicitly states that one observed value is inaccurate and asks for an estimate of the correct value, use the fitted straight line to estimate the corresponding transformed Y-value at the given X, then reverse the transformation.

For example, if Y = y/x and the fitted graph suggests Y = 7.4 when x = 8, then y ≈ xY = 8(7.4) = 59.2.

The estimate comes from the model represented by the line, not from arbitrarily averaging nearby y-values.

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12 · Linearisation is also a model test

If the chosen transformation is correct and the model is a good approximation over the observed range, the transformed points should be approximately linear.

If the points show systematic curvature, one of several things may be wrong:

• the proposed model family is inappropriate;
• the transformation was derived incorrectly;
• a parameter or variable has been misunderstood;
• the relationship changes across the measured range;
• the data quality is poor.

A straight transformed graph is therefore evidence in favour of the model, not merely a convenient shape.

Straight does not mean exact

Experimental points rarely fall on a mathematically perfect line. “Approximately straight” supports an approximate model. The fitted constants inherit experimental uncertainty. A classroom answer may report them to a sensible number of significant figures rather than pretending the graph has produced exact constants.

This is a useful distinction between pure algebra and modelling. Algebraic transformations may be exact; measurements and parameter estimates may not be.

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13 · Residual thinking: what the line fails to explain

The current syllabus does not require a formal statistics course in residual analysis here, but a modern learner benefits from the concept. For each transformed data point, imagine comparing the observed Y-value with the Y-value predicted by the fitted line. The difference is a residual.

If residuals are small and scattered without a pattern, the straight-line model may be reasonable. If they are systematically positive in one region and negative in another, the transformed relationship may still contain curvature.

This strengthens the old textbook idea of judging a line of best fit. We are not merely drawing a visually pleasing line; we are checking whether a linear representation explains the transformed data without a systematic leftover structure.

Why this matters in the age of graphing software

Modern software can fit a nonlinear curve directly. That does not make Linear Law obsolete. Linearisation teaches the learner how the model is built and how the constants map into visible graph features. It also allows a human to inspect model adequacy rather than simply accepting a regression output because a device produced it.

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14 · Interpolation and extrapolation

Once a model is recovered, it can be used to estimate unknown values.

• Interpolation: estimate inside the observed x-range.
• Extrapolation: estimate outside the observed x-range.

Interpolation is usually more defensible because the model is being used where it has observational support. Extrapolation assumes the same relationship continues beyond the data. That may be reasonable for a short distance and unreliable for a long one.

Worked Example 9 · Return through the transformation

A fitted power-law model is y = 3.2x^1.4. Estimate y when x = 5.

y = 3.2(5)^1.4 ≈ 30.5.

If x = 5 lies inside the experimental range, this is interpolation. If the observed x-values ended at 2, it is extrapolation and should be treated with more caution.

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15 · Reverse transformations must use the correct base

One of the most common high-cost errors is recovering a parameter with the wrong inverse operation.

• If c = ln a, then a = e^c.
• If c = log₁₀a, then a = 10^c.
• If m = log₁₀b, then b = 10^m.
• If m = ln b, then b = e^m.

A student who writes “a = c” has forgotten the return path. A student who writes a = 10^c after using ln has remembered that a return is needed but used the wrong inverse base.

Write the transformed equation explicitly before interpreting the graph. The equation itself tells you what inverse operation is required.

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16 · A compact decision tree

1. Identify the model family. Power, exponential or another rearrangeable relation?
2. Choose a transformation. Logarithm, reciprocal, division, multiplication, subtraction of a known nonlinear term?
3. Rearrange completely. The final equation must look like Y = mX + c.
4. Name the axes. What is plotted against what?
5. Read m and c. Use appropriate points on the line or given graph information.
6. Map m and c to the original constants.
7. Reverse transformations. Antilog, exponential, reciprocal or algebraic rearrangement as required.
8. Rebuild the original model.
9. Check at least one known condition or data point.
10. Interpret model quality. Is the transformed graph plausibly linear over the observed range?

Do not stop at the straight line. The answer belongs back in the original nonlinear world.

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17 · The Linear Law error map

“Weak at Linear Law” can mean many different things. The correct repair depends on the first failed stage.

• Target failure: the learner manipulates the equation without aiming for Y = mX + c.
• Log-law failure: log(axⁿ) is expanded incorrectly.
• Axis failure: a correct equation is paired with the wrong horizontal or vertical variable.
• Gradient-intercept swap: m and c are confused.
• Parameter mapping failure: log b is mistaken for b.
• Base failure: antilog base 10 is used after natural logarithms.
• Transformation arithmetic failure: y/x, 1/x or log values are calculated inaccurately in a data table.
• Graph reading failure: gradient is taken from two nearby points or from raw data rather than the fitted line.
• Return-path failure: parameters are recovered but never substituted into the original model.
• Model-fidelity failure: a visibly curved transformed plot is treated as acceptable because a straight line was expected.
• Anomaly failure: an outlying point is deleted automatically instead of investigated.
• Scope failure: the learner memorises a huge catalogue of transformations without mastering the explicit K341 power and exponential forms first.

The first weak-link diagnostic

1. Can the learner derive a straight-line form from y = axⁿ?
2. Can the learner state the correct axes?
3. Can the learner identify m and c?
4. Can the learner recover a and n?
5. Can the learner repeat the process for y = kb^x?
6. Can the learner recover parameters from two transformed graph points?
7. Can the learner verify the recovered model?

Stop at the first unstable step. A learner who cannot state X and Y does not need a harder graph. A learner who gets the graph right but sets b equal to the gradient needs parameter-return practice.

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18 · SEC examination transfer

A complete G3 preparation sequence should not only ask students to “convert to linear form”. It should vary the direction and missing information.

• Given the nonlinear model, determine the axes.
• Given a straight-line equation, reconstruct the nonlinear model.
• Given two points on the transformed graph, find the unknown constants.
• Given a table, calculate transformed values and plot the graph.
• Given one anomalous observation, estimate a corrected value from the fitted model.
• Given two possible transformations, choose the clearer one.
• Given a transformed graph that curves, explain what that suggests about the proposed model.
• Given a recovered model, estimate an unobserved value and distinguish interpolation from extrapolation.

These variations train the actual objective: representation selection and parameter recovery. A student who has only practised one fixed template is vulnerable as soon as the question reverses the direction.

19 · Original guided practice

The following questions are original to this classroom guide. They move from direct transformation to full model recovery and interpretation.

A · Match to Y = mX + c

1. Transform y = axⁿ to linear form using logarithms. State X, Y, m and c.
2. Transform y = kb^x to linear form using common logarithms. State X, Y, m and c.
3. Transform y = ae^bx to linear form using natural logarithms.
4. Transform y = a + b/x into straight-line form without logarithms.
5. Transform y = ax² + bx into straight-line form by dividing through by x.
6. Transform y = x³ + ax + b into straight-line form.
7. Transform xy = ax² + b into straight-line form.

B · Read the transformed graph

8. For y = axⁿ, a graph of log y against log x has gradient 2.3 and intercept 0.6990. Find a and n.
9. For y = kb^x, a graph of log y against x has gradient 0.3010 and intercept 1.4771. Find k and b.
10. For y = ae^bx, a graph of ln y against x has gradient −0.25 and intercept ln20. Find a and b.
11. A graph of y/x against x is a straight line with gradient 3 and intercept −4. Find a and b if y = ax² + bx.
12. A graph of y against 1/x has gradient 7 and intercept 2. Find a and b if y = a + b/x.

C · Use two transformed points

13. For y = ae^bx, the graph of ln y against x passes through (2,3) and (6,5). Find a and b.
14. For y = axⁿ, the graph of log y against log x passes through (1,1.2) and (3,4.0). Find n and express a in terms of a power of 10.
15. For y = kb^x, the graph of log₁₀y against x passes through (0,0.6021) and (4,2.5106). Find k and b.
16. For y = a + b/x, the graph of y against 1/x passes through (1/2,7) and (1/6,3). Find a and b.

D · Transform data

17. The following values are believed to satisfy y = ax² + bx: x = 2,4,6,8 and y = 12,40,84,144. Calculate y/x and recover a and b.
18. The values x = 1,2,4,8 and y = 6,17,53,197 are believed to satisfy y = a + b/x. Test this claim qualitatively by considering what graph should be straight and whether the transformed values appear consistent.
19. The values x = 1,2,4,8 and y = 6,18,54,162 are believed to satisfy y = kb^x. State the transformation required and explain whether the raw pattern itself already gives a useful clue.

E · Model checking and anomalies

20. A transformed data set should lie on a straight line. Five points are close to the line and one is far away. Give four possible explanations for the outlying point.
21. Explain why an outlying transformed point should not automatically be deleted.
22. A fitted graph of y/x against x predicts Y = 7.4 at x = 8. Estimate y.
23. Explain why using two far-apart points on a fitted line usually gives a more stable gradient estimate than two very close points.
24. A transformed plot bends systematically upward. What does this suggest about the chosen model or transformation?

F · Mixed SEC reasoning

25. For y = axⁿ, a student plots log x against log y instead of log y against log x. Explain what changes.
26. A student correctly obtains ln y = 0.8x + 1.5 from y = ae^bx, then writes a = 1.5 and b = 0.8. Identify the error and correct a.
27. A student uses common logs to obtain log y = 0.4x + 0.7 for y = kb^x. Write k and b exactly in terms of powers of 10.
28. A straight-line graph of log y against log x has zero gradient. What does that imply about n in y = axⁿ, and what does the original model become?
29. A graph of ln y against x has negative gradient. What does that imply about b in y = ae^bx?
30. Explain why y = axⁿ cannot be linearised with real logarithms using data points for which x ≤ 0 if n and the model do not guarantee appropriate real values.
31. Two different algebraic transformations both produce straight-line form. Give two criteria for choosing between them.
32. A recovered exponential model fits the observed data from x = 1 to x = 5. Explain why predicting x = 100 should be treated cautiously.
33. Explain the difference between an exact algebraic transformation and an approximate fitted straight line.
34. State the full return path from nonlinear model to transformed graph and back to verified model.

20 · Worked answers and reasoning checkpoints

1. log y = n log x + log a. Y = log y, X = log x, m = n, c = log a.
2. log y = x log b + log k. Y = log y, X = x, m = log b, c = log k.
3. ln y = bx + ln a. Y = ln y, X = x, m = b, c = ln a.
4. y = b(1/x) + a. Y = y, X = 1/x, m = b, c = a.
5. y/x = ax + b. Y = y/x, X = x, m = a, c = b.
6. y − x³ = ax + b. Y = y − x³, X = x, m = a, c = b.
7. xy = a(x²) + b. Y = xy, X = x², m = a, c = b.
8. n = 2.3. log a = 0.6990, so a ≈ 5. Model y ≈ 5x^2.3.
9. log b = 0.3010, so b ≈ 2. log k = 1.4771, so k ≈ 30.
10. a = 20, b = −0.25.
11. y/x = 3x − 4, so a = 3, b = −4.
12. y = 7(1/x) + 2, so a = 2, b = 7.
13. Gradient b = (5 − 3)/(6 − 2) = 1/2. Equation ln y = 0.5x + c. Using (2,3): c = 2. Thus ln a = 2, so a = e², b = 1/2.
14. n = (4.0 − 1.2)/(3 − 1) = 2.8/2 = 1.4. At X = 1, 1.2 = 1.4(1) + c, so c = −0.2. Thus a = 10^−0.2.
15. At x = 0, log k = 0.6021, so k ≈ 4. Gradient = (2.5106 − 0.6021)/4 ≈ 0.4771, so b ≈ 3. k ≈ 4, b ≈ 3.
16. Gradient b = (7 − 3)/(1/2 − 1/6) = 4/(1/3) = 12. Use 7 = 12(1/2) + a → a = 1. a = 1, b = 12.
17. y/x values are 6,10,14,18. They follow Y = 2X + 2, so a = 2, b = 2.
18. For y = a + b/x, plot y against 1/x. The raw pairs do not obviously approach a constant intercept pattern in a simple way; the transformed table is needed. This is a reminder that model checking should follow the correct transformation rather than visual guessing from the raw table alone.
19. For y = kb^x, plot log y against x. The values triple when x increases by 1 only across the first steps listed in powers-of-two spacing the x-values do not increase uniformly, so the raw sequence must be interpreted carefully. The transformation, not visual pattern alone, is the reliable test.
20. Possible causes include measurement error, recording error, transformed-value calculation error, changed experimental conditions, random variation, or model breakdown.
21. Because an outlier may contain real information about the experiment or model. It should be investigated, not deleted merely to make the graph straighter.
22. Y = y/x = 7.4 and x = 8, so y ≈ 59.2.
23. A fixed coordinate-reading error is a smaller fraction of a large rise/run than of a small rise/run, so a wide triangle generally gives a more stable slope estimate.
24. Systematic curvature suggests the model family may be unsuitable, the transformation may be wrong, or the relationship may change across the range.
25. Reversing the axes changes the line. If log y = n log x + log a, then log x = (1/n)log y − (log a)/n when n ≠ 0. The new gradient is 1/n, not n, and the intercept has a different parameter meaning.
26. c = ln a, so a is not 1.5. a = e^1.5; b = 0.8 is correct.
27. k = 10^0.7, b = 10^0.4.
28. Zero gradient means n = 0. Then y = ax⁰ = a, a constant relationship for admissible x.
29. The gradient equals b, so b < 0; the model shows exponential decay.
30. Real logarithms require positive arguments. A log–log transformation needs values for which x and y are positive. Data outside that domain cannot simply be inserted into the same real logarithmic graph.
31. Choose the form that makes transformed values easier to calculate, gives clearer parameter recovery, uses stable scales, minimises error amplification, or matches the supplied graph/data more naturally.
32. That is long-range extrapolation. The model has observational support only over the measured range, and the underlying process may change before x = 100.
33. The algebraic transformation can be exact: every valid original pair maps exactly to the transformed relationship. The fitted straight line is approximate when the data contain measurement variation or the model is only approximate.
34. Original model → algebraic transformation → define X,Y → straight-line form → plot/read m,c → map m,c to original parameters → reverse transforms → reconstruct original model → verify against data/conditions → assess model fit.

Checking Question 19: the deliberately awkward raw data are a reminder that one should not infer an exponential law merely because some y-values appear to multiply. The x-spacing matters. Linear Law formalises the check by asking whether the correctly transformed points align.

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21 · A two-week revision architecture

Linear Law disappears quickly when students revise by memorising a table of transformations. Retrieval must instead rebuild the transformation from the model.

1. Day 1: derive Y = mX + c for y = axⁿ and y = kb^x.
2. Day 2: write axes, gradient and intercept meanings from memory.
3. Day 4: reverse transformed graphs into original models.
4. Day 6: practise natural exponential y = ae^bx.
5. Day 8: add two non-logarithmic transformations from the legacy chapter.
6. Day 10: transform short data tables and recover constants.
7. Day 12: identify anomalous points and estimate corrected values where the question justifies it.
8. Day 14: complete a mixed unlabelled set requiring the learner to choose the transformation independently.

The six-minute retrieval test

• Linearise y = axⁿ.
• Linearise y = kb^x.
• State the axes for both graphs.
• Explain the difference between gradient and the original parameter b in y = kb^x.
• State one non-logarithmic linearisation.
• Describe the return path after m and c are read from the graph.

22 · For teachers: teach backwards from the target

The most productive classroom question is often not “What should we do to this equation?” but “What would have to be true for this to look like Y = mX + c?”

This shifts manipulation from trial-and-error into goal-directed algebra. For y = axⁿ, the exponent n needs to become a multiplier; logarithms do that. For y = ax² + bx, the common factor x can be removed by division, exposing ax + b. For y = x³ + ax + b, the known nonlinear term x³ can be subtracted.

Students should say what each transformation is accomplishing. That sentence-level explanation often reveals whether they understand the method or are reproducing an old example by resemblance.

Do not teach a giant transformation lookup table first

A lookup table can help with later revision, but it is a poor starting point. It encourages students to match surface patterns rather than derive. Begin with the two explicit SEC families, then add a small number of algebraic variants so the learner understands the general principle.

Use graph imperfection as a teaching asset

Do not make every data set perfect. Realistic scatter, one suspicious point and sensible rounding create opportunities to discuss model fit, measurement error and why a fitted line is an estimate rather than an oracle.

The old textbook’s experimental framing remains excellent here. Modernise the interpretation, not the underlying reason for teaching it.

23 · For parents: what progress sounds like

A student who has memorised Linear Law may say, “I think I have to take logs.” A student who understands the chapter begins to say:

• “I want Y = mX + c, so I need to make the exponent become a multiplier.”
• “For y = axⁿ, I should plot log y against log x.”
• “The gradient gives n directly, but the intercept gives log a, so I still have to recover a.”
• “For y = kb^x, the gradient is log b, not b.”
• “The points do not look straight after the correct transformation, so the model may not fit.”
• “That point is an outlier, but I need a reason before treating it as wrong.”
• “I have the constants now; I should put them back into the original model and check.”

This language is evidence that the learner understands the transformation-and-return cycle rather than only the graphing procedure.

24 · What carries forward

Linear Law sits at a useful crossroads. It uses logarithms from the previous algebra chapter, gradient and intercept from coordinate geometry, model interpretation from functions, and verification habits that will matter throughout calculus.

The old textbook moves next into Trigonometric Functions and Equations, followed by Trigonometric Identities and Formulae. The 2027 SEC structure combines much of that material under a single trigonometric-functions, identities and equations strand for both G2 and G3 Additional Mathematics. The next classroom chapter will therefore reunify the old Chapters 8 and 9 where the current syllabus does so, while keeping the longform textbook walkthrough readable and staged.

The conversion principle remains the same: preserve strong teaching sequences, remove obsolete ownership boundaries, and let the current syllabus decide which ideas belong together.

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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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