Additional Mathematics Synthesis Guide 18: Linear Law Transformations, Straight-Line Graphs, Parameter Recovery and Model Checking

Original relationship → transform → Y=mX+c → read gradient/intercept → recover parameters → return to original model.

1. Linear law is a change of representation

The basic straight-line equation is

Y=mX+c.

In a linear-law problem, X and Y may not be the original variables x and y. They might be ln x, ln y, 1/x, xy, y/x or another transformed quantity.

The key question is therefore not merely “What is the gradient?” but “What quantity is on each axis, and how does that gradient map back to the original parameters?”

2. Power law: y=Axⁿ

Suppose x>0 and y>0, with

y=Axⁿ, A>0.

Take natural logarithms:

ln y = n ln x + ln A.

Define

• Y=ln y
• X=ln x

Then Y=nX+ln A. Therefore a plot of ln y against ln x has

• gradient n
• vertical intercept ln A

Recover A by exponentiating the intercept: A=e^c.

3. Worked power-law recovery

A graph of ln y against ln x is a straight line with gradient 1.5 and vertical intercept ln4. Find the original model.

From the gradient, n=1.5=3/2. From the intercept, A=4. Hence

y=4x^3/2.

If x=9, the model predicts y=4(27)=108.

Do not write A=ln4. The intercept is ln A, not A itself.

4. Base-10 logarithms work too

Using common logarithms instead gives

log y=n log x+log A.

The gradient remains n. The intercept becomes log A, so A=10^c.

Natural logs and common logs both linearise a power law. The error is mixing conventions: if the graph uses log base 10, do not recover A with e^c.

5. Exponential law: y=Ae^kx

For y=Ae^kx with A>0, take natural logarithms:

ln y=kx+ln A.

A plot of ln y against x is straight with

• gradient k
• intercept ln A

This differs from the power law because only y is logged. The horizontal axis remains x.

6. Worked exponential recovery

A plot of ln y against x has gradient −0.25 and vertical intercept ln80. Then

y=80e^−0.25x.

The transformed graph decreases linearly, while the original positive quantity decays exponentially. The slope −0.25 belongs to ln y versus x, not to y versus x.

At x=4, y=80e⁻¹=80/e.

7. Exponential base b: y=Ab^x

For A>0 and b>0,

ln y=x ln b+ln A.

Therefore on a graph of ln y against x:

• gradient = ln b
• intercept = ln A

Recover b as e^m, not m itself.

If m=ln1.2, then b=1.2. Equal increases of 1 in x multiply y by 1.2 under the model.

8. Reciprocal law: y=a+b/x

Suppose

y=a+b/x.

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

Y=bX+a.

A graph of y against 1/x has gradient b and vertical intercept a.

This transformation does not require logarithms. The useful structure is already affine in 1/x.

9. Worked reciprocal recovery from two transformed points

A graph of y against 1/x passes through the transformed points (0.2,7) and (0.5,13). Find a and b in y=a+b/x.

The transformed gradient is

b=(13−7)/(0.5−0.2)=6/0.3=20.

Using (0.2,7):

7=20(0.2)+a, so a=3.

Thus

y=3+20/x.

The horizontal transformed coordinate 0.2 means 1/x=0.2, so the original x-value there is 5. Confusing these two coordinate systems is a common error.

10. A product transformation: xy=ax+b

Suppose a relationship can be rearranged to

xy=ax+b.

Let Y=xy and X=x. Then Y=aX+b. A graph of xy against x has gradient a and intercept b.

This kind of problem tests whether the learner can see the whole left-hand product as the transformed dependent variable. Plotting y against x instead would generally not give the required straight line.

11. Choosing the axes from a target form

Given

y=(a/x)+bx,

multiply by x:

xy=a+bx².

Now define Y=xy and X=x². Then

Y=bX+a.

So a plot of xy against x² is linear, with gradient b and intercept a.

The transformation was not guessed from a list. It was engineered by rearranging the original relation into Y=mX+c.

12. Worked transformation design

Suppose

y=(p+qx²)/x.

Multiply by x:

xy=p+qx².

Thus plot xy vertically against x² horizontally. The line has gradient q and intercept p.

If the transformed line passes through (1,5) and (4,14), then q=(14−5)/(4−1)=3 and p=2. The original model is

y=(2+3x²)/x=2/x+3x.

13. Intercepts may be transformed constants, not original y-values

For ln y=n ln x+ln A, the transformed vertical intercept is ln A. It is not automatically a value of the original y-axis because the vertical axis is ln y.

When ln x=0, the original x is 1. At that point ln y=ln A, so y=A. This is a legitimate original interpretation, but it comes through the inverse transformation.

Similarly, if the horizontal axis is 1/x, the transformed x-intercept X=0 corresponds to the limit 1/x→0, not to the original value x=0. Axis labels change what coordinates mean.

14. Recovering parameters from gradient and one point

Suppose y=Axⁿ, and a log-log graph has gradient 2.5. It also passes through the transformed point (ln2,ln24).

Then n=2.5. Substitute into

ln24=2.5ln2+ln A.

Hence

ln A=ln24−ln(2^2.5)

so

A=24/2^5/2=3√2.

The model is y=3√2 x^5/2.

15. Two original data points can determine a power law directly

Suppose y=Axⁿ and the points (2,12) and (8,96) lie on the model.

Divide the equations:

96/12=(8/2)ⁿ.

Thus 8=4ⁿ, so n=3/2. Then

12=A·2^3/2, so A=3√2.

The same result would appear as gradient 3/2 on a log-log graph. Linearisation is a representation that makes the parameter relation visually accessible; it is not the only algebraic route.

16. Straight transformed data support, but do not prove, the model

If measured points lie approximately on a straight line after the proposed transformation, that is evidence that the model is useful over the observed range. It is not a proof that the exact physical law is y=Axⁿ or Ae^kx in every circumstance.

Real data contain measurement noise, omitted variables and changing conditions. A linearised plot should therefore be treated as part of model checking: inspect the scatter, systematic curvature, unusual points and the range over which data were collected.

The mathematics tells us what pattern the model predicts. Evidence tells us how well the observed situation follows that pattern.

17. Curvature in transformed space is diagnostic

If a proposed exponential model predicts ln y against x should be straight but the transformed data show systematic curvature, then at least one assumption is questionable: the model form, the data range, a parameter constancy assumption, or the measurements themselves.

Random-looking scatter around a line and smooth systematic curvature are different evidence patterns. Both create imperfect straightness, but they suggest different follow-up questions.

Do not repair curvature merely by drawing a more convenient straight line through it. The purpose of transformation is to test structure, not conceal disagreement.

18. Units and logarithms require care

In school algebra, expressions such as ln y are often handled formally. In physical modelling, taking logarithms of a dimensional quantity is interpreted through a dimensionless ratio relative to a reference unit or scale.

For examination questions, follow the stated transformed axes and expected conventions. For real modelling, keep units explicit and avoid treating a change of measurement unit as though it should alter the underlying physical relationship.

A parameter’s numerical intercept may change under a unit conversion even when the underlying phenomenon has not changed. Interpretation must return to the original variables and units.

19. The gradient depends on which variable is vertical

A graph of Y against X has gradient ΔY/ΔX. Reversing the axes generally gives reciprocal gradient only when the relation is exactly linear and the original gradient is non-zero.

In a parameter-recovery problem, swapping the axes can therefore change the formula mapping completely. “Plot ln y against ln x” is not interchangeable with “plot ln x against ln y”.

Read “A against B” as A vertical, B horizontal unless a graph explicitly states otherwise.

20. A model can have a valid transformed line but an invalid input

For a power-law log transformation, x and y must be positive if real logarithms are used. A transformed straight-line equation does not authorise negative or zero original inputs that the logarithm could not represent.

For reciprocal transformations, x=0 remains excluded. For square-root transformations, the original radicand conditions remain. The transformation is a new coordinate system for legal data, not a device for expanding the original domain.

21. Common failure patterns

• Reading ln A as A.
• Plotting the original variable where the transformed variable was required.
• Using e^c to undo a base-10 logarithmic intercept.
• Confusing the transformed coordinate ln2 with the original coordinate 2.
• Swapping vertical and horizontal axes and retaining the old gradient interpretation.
• Forgetting positivity conditions for logarithmic transformations.
• Assuming a roughly straight transformed plot proves a universal law.
• Ignoring systematic curvature because a straight line can still be drawn.

22. A reliable linear-law routine

1. Write the original model clearly.
2. Algebraically rearrange toward Y=mX+c.
3. Define X and Y explicitly.
4. Label transformed axes before using gradient or intercept.
5. Map m and c back to the original parameters.
6. Undo logarithms or reciprocals with the correct inverse operation.
7. Check recovered parameters in one original data point.
8. Return to original units and discuss model validity over the observed range.

23. Practice set

1. Linearise y=Axⁿ using natural logarithms.
2. On a graph of ln y against ln x, what do gradient 2 and intercept ln5 imply?
3. Linearise y=Ae^kx.
4. A graph of ln y against x has gradient −0.6 and intercept ln30. Find the model.
5. For y=Ab^x, what does the gradient of ln y against x equal?
6. If that gradient is ln1.4, find b.
7. Linearise y=a+b/x by choosing X and Y.
8. A y-against-1/x graph has gradient 12 and intercept 4. Find the original model.
9. Rewrite y=(p+qx²)/x in straight-line form and state suitable axes.
10. A graph of xy against x² has gradient 5 and intercept −2. Find p and q.
11. If log₁₀y is plotted against log₁₀x with intercept 2, what is A in y=Axⁿ?
12. If ln y is plotted against ln x with intercept 2, what is A?
13. Two points on y=Axⁿ are (1,3) and (4,24). Find A and n.
14. For y=Ae^kx, Q(0)=50 and Q(2)=100. Find k.
15. Explain why plotting ln y against x would not be the standard power-law linearisation of y=Axⁿ.
16. What does systematic curvature in the transformed graph suggest?
17. Why should x>0 be recorded when using ln x?
18. What original x-value corresponds to transformed coordinate X=ln8?
19. What original x-value corresponds to X=1/x=0.25?
20. Give one independent way to verify recovered parameters.

Answers

1. ln y=n ln x+ln A.
2. n=2 and A=5, so y=5x².
3. ln y=kx+ln A.
4. y=30e^−0.6x.
5. ln b.
6. 1.4.
7. Y=y, X=1/x, giving Y=bX+a.
8. y=4+12/x.
9. xy=p+qx²; plot Y=xy against X=x².
10. p=−2, q=5.
11. A=10²=100.
12. A=e².
13. A=3. Then 24/3=4ⁿ, so 8=4ⁿ and n=3/2.
14. 100=50e^2k, so k=(ln2)/2.
15. Because ln y=ln A+n ln x is linear in ln x, not generally in x.
16. The proposed model or constant-parameter assumption may not fit the observed range; investigate rather than forcing a line.
17. Because real ln x is defined only for positive x.
18. x=8.
19. x=4.
20. Substitute a recovered parameter set into one original data point or compare a model prediction with an unused data point.

24. What mastery looks like

Mastery means the learner can design the transformation rather than only recognise a memorised pair of axes, can distinguish transformed coordinates from original values, can recover parameters from gradient and intercept with the correct inverse operation, and can treat transformed straightness as evidence about a model rather than proof by appearance.

The transfer test is to supply an unfamiliar relationship that can be rearranged into Y=mX+c without naming the axes. If the learner can create suitable transformed variables, recover parameters and return to the original model with restrictions intact, linear law has become a general representation skill.

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• Guide 17: R-Form Trigonometry, Amplitude, Phase, Maxima and Modelling
• Guide 19: Trigonometric Calculus — Differentiation, Integration, Stationary Points and Exact Areas
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