For Integrated Programme students: Your current school materials, teacher instructions, and assessment scope take precedence because IP topic sequence and depth vary by school. This is an Eclat IP guide, not the O-Level / SEC G3 exam-track guide.
How this chapter applies
Eclat core: transforming power and exponential relationships into straight-line form, reading gradient and intercept, and recovering unknown constants form the main route.
School-sensitive extension: reciprocal, mixed, or log-linear models beyond the named K341 families may be school-sensitive depth.
2027 national comparison: K341 Topic G2.5 explicitly requires transforming relationships including y = ax^n and y = kb^x to linear form to determine unknown constants from a straight-line graph.
Check your school: confirm permitted axis transformations, plotting precision, and whether regression tools are allowed.
Q: What does IP AMaths Notes (Upper Sec, Year 3-4): 03) Linear Law cover? A: Transform non-linear relationships into straight lines for graph paper work and regression checks in IP AMaths.
Linear-law questions ask you to recast a model so plotting gives a straight line. Identify the transformation, compute plotting coordinates, and interpret the resulting gradient and intercept.
Keep the full topic roadmap handy via our IP Maths tuition hub so you can jump into related drills, quizzes, or diagnostics as you move through these notes.
The core idea is simple: Linear Law turns a curved model into a straight-line graph.
Use it as a working check: Decide what to plot on each axis before calculating values. The gradient and intercept must translate back to the original constants.
Then go one layer deeper: Example: for a power model y = A x^n, plot log(y) against log(x); the gradient gives n and the intercept gives log(A).
Choosing the plotting axes
Before making a table of values, decide what quantity should become the straight-line vertical axis and what should become the horizontal axis. This prevents the common mistake of plotting the original y against the original x when the model is not linear.
Model form
Plot vertical axis
Plot horizontal axis
What gradient and intercept mean
y=Abx
logy
x
Gradient is logb; intercept is logA.
y=Axn
logy
logx
Gradient is n
y=a+bx1
y1
y=a+blnx
y
lnx
Gradient is b
y=Ae−kx
lny
x
Gradient is −k
Common trap: The intercept is often a logged constant, not the constant itself. If the intercept is logA, convert back with A=10intercept before stating the model.
Gradient-intercept decoding checkpoint
After drawing the straight-line graph, do not stop at the gradient and intercept. Decode them back into the constants in the original model.
Transformed equation:
Y = mX + c
Graph tells you:
gradient = m
vertical intercept = c
Question usually wants:
original constants such as A, b, n, a, or k
Original model
Straight-line form
What the graph gives
What to report
y=Abx
logy=xlogb+logA
gradient m=logb, intercept c=logA
b=10m, A=10c if base-10 logs are used.
y=Axn
logy=nlogx+logA
y=Ae−kx
lny=−kx+lnA
Misconception check: the intercept is not always the original constant. It is the constant in the transformed straight-line equation, so convert it back before writing the final model.
Log-base consistency checkpoint
Once you choose a log base for a transformed graph, keep that base all the way to the final constants. Mixing log and ln changes the gradient and intercept values, even when the straight-line idea is the same.
If the graph uses
Intercept c means
Convert back with
Common trap
log10y against x for y=Abx
c=log10A
A=10c
Using ec because the model looks exponential.
lny against x for y=Ae−kx
c=lnA
log10y against log10x for y=Axn
Worked check: if a straight-line graph of lny against x has intercept 1.20, then lnA=1.20, so A=e1.20≈3.32. If the graph used log10y instead, the same intercept value would give A=101.20≈15.8.
Misconception check: the symbol on the graph axis controls the undo step. Do not decide between 10c and ec from the original word "exponential" alone.
1 Standard transformations
Exponential form y=Abx: take logs to obtain logy=logA+xlogb.
Power form y=Axn: take logs to produce logy=logA+nlogx.
Reciprocal form y=ax+b1: rearrange to y1=ax+b
Mixed models (e.g. y=xnA) combine the above manipulations.
2 Worked example - Power model
Given y=kx1.5, rewrite in linear-law form and determine k from the data (x,y)=(4,32), (25,500).
Take base-10 logs: log10y=log10k+1.5log10x.
Let Y=log10y, X=log10x
Compute: X1=log104=0.6021, Y1=log1032=1.5051
For the second point: X2=log1025=1.3979, Y2=log10500=2.6990
Gradient of straight line: m=1.3979−0.60212.6990−1.5051=1.500, matching the exponent in the model.
Intercept: C=Y1−mX1=1.5051−1.500×0.6021=0.60195
Hence k=100.60195≈4.00 (3 s.f.).
3 Worked example - Reciprocal model
A dataset follows y=a+bx1. Show how to plot a straight line and determine a, b using the points (1,0.42), (3,0.21).
Rearrange: y1=a+bx.
Define Y=y1; plotting Y against x gives intercept a and gradient b.
Compute values: for x=1, Y1=0.421=2.381
Gradient: b=3−14.762−2.381=1.1905.
Intercept: a=Y1−b=2.381−1.1905=1.1905.
4 Worked example - Log-linear model
Suppose sensor readings obey y=a+blnx. Given data points (2,4.1), (5,6.7), (9,8.4), estimate a and b using a straight-line plot.
Transform the x-values: compute X=lnx. This gives X1=0.6931, X2=1.6094, X3=2.1972 (4 d.p.).
Plot y against X; the model becomes y=a+bX so the gradient equals b.
Use the first and last points to estimate the gradient: b≈2.1972−0.69318.4−4.1=2.86.
Find the intercept with any point: a≈4.1−2.86×0.6931=2.12.
State the fitted model: y≈2.12+2.86lnx. Always mention plotting accuracy limits when quoting constants.
Answer: take a≈2.12 and b≈2.86.
5 Practice Quiz
Check that you can choose the right straight-line transform, extract gradients/intercepts, and translate them back into model parameters quickly.
6 Try this
Data believed to satisfy y=Ae−kx. Outline how you would estimate A and k from a semi-log plot, then test with two sample points of your choice.
