Linear Law — Straight-Line Form

Q: Can I avoid using logs for all non-linear relationships?

Only for simple polynomial forms like , which can be rearranged directly. Power and exponential relationships require logs to eliminate exponents on variables.

CIE IGCSE Additional Mathematics· 7.4· 12 min read

1. Transforming Polynomial Relationships★★☆☆☆⏱ 3 min

Many non-linear relationships can be converted to straight-line form with simple algebraic rearrangement, no logarithms required. This applies to polynomial forms where x and y are raised to fixed constants, like $y^{2} = A x^{3} + B$ .

📘 Definition

Straight-Line Form Transformation

The process of rearranging a non-linear relationship to match $Y = m X + c$ , where Y and X are functions of the original variables x and y, m is the gradient, and c is the Y-intercept.

📐 Worked Example

Transform the relationship $y^{2} = 5 x^{3} + 2$ into $Y = m X + c$ form, identifying Y, X, m and c.

1
Compare the given form to the target straight line structure. The dependent term is $y^{2}$ , and the independent term is $x^{3}$ .
2
Define transformed variables: $Y = y^{2}$ and $X = x^{3}$ .
3
Substitute into the original equation: $Y = 5 X + 2$ .
4
Identify values: Gradient $m = 5$ , intercept $c = 2$ .

Exam tip:

Always clearly label your transformed Y and X variables in exam answers to avoid losing method marks.

2. Linearizing Power Relationships ($y = Ax^n$)★★★☆☆⏱ 3 min

Power relationships have x raised to an unknown constant n, so we use logarithm rules to eliminate the exponent and create a linear form. Taking log base 10 of both sides converts the product and power into additive terms.

📐 Worked Example

Transform $y = A x^{n}$ into straight line form, and state how to find A and n from the transformed graph.

1
Take log₁₀ of both sides of the equation: $l g y = l g (A x^{n})$ .
2
Apply logarithm product rule: $l g y = l g A + l g x^{n}$ .
3
Apply logarithm power rule: $l g y = n l g x + l g A$ .
4
Match to $Y = m X + c$ : $Y = l g y$ , $X = l g x$ , gradient $m = n$ , intercept $c = l g A$ .
5
Calculate constants: n equals the gradient of the graph, $A = 1 0^{c}$ where c is the Y-intercept.

3. Linearizing Exponential Relationships ($y = Ab^x$)★★★☆☆⏱ 3 min

Exponential relationships have x in the exponent, so we also use logarithms to linearize these forms. The process is nearly identical to power relationships, but the transformed X variable is the original x, not a log function of x.

📐 Worked Example

Transform $y = A b^{x}$ into straight line form, and state how to find A and b from the transformed graph.

1
Take log₁₀ of both sides: $l g y = l g (A b^{x})$ .
2
Apply product rule: $l g y = l g A + l g b^{x}$ .
3
Apply power rule: $l g y = (l g b) x + l g A$ .
4
Match to $Y = m X + c$ : $Y = l g y$ , $X = x$ , gradient $m = l g b$ , intercept $c = l g A$ .
5
Calculate constants: $b = 1 0^{m}$ , $A = 1 0^{c}$ .

✓ Quick check

For the relationship $y = 3 (2)^{x}$ , what is the gradient of the transformed graph of $l g y$ against x?
Reveal answer
$\lg 2 \approx 0.3010$ —
Correct! The gradient equals $l g b$ where b is the base of the exponential, here b=2.

4. Solving Linear Law Exam Problems★★★★☆⏱ 4 min

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Most exam questions will give you experimental x and y values, or a plot of the transformed straight line, and ask you to calculate unknown constants. Always show all transformation steps to earn full method marks.

📐 Worked Example

A plot of $l g y$ against $l g x$ gives a straight line with gradient 2 and Y-intercept 0.69897. Find the values of A and n for the relationship $y = A x^{n}$ .

1
Recall the transformed form for power relationships: $l g y = n l g x + l g A$ .
2
Match gradient to n: $n = 2$ .
3
Match intercept to $l g A$ : $l g A = 0.69897$ .
4
Calculate A: $A = 1 0^{0.69897} = 5$ .
5
Final relationship: $y = 5 x^{2}$ .

5. Linearizing $e$ and $\ln$ Relationships★★★★☆⏱ 4 min

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The natural exponential $e$ and natural logarithm $ln$ are part of the 0606 syllabus, and the linear-law examples in section 7.4 include forms such as $e^{2 y} = A x^{2} + B$ and $y^{3} = A ln x + B$ . Handle these exactly like the earlier cases: choose transformed variables Y and X so the equation becomes $Y = m X + c$ , then read the unknown constants from the gradient and intercept.

📐 Worked Example

The variables x and y satisfy $e^{2 y} = A x^{2} + B$ , where A and B are constants. When $e^{2 y}$ is plotted against $x^{2}$ , a straight line is obtained passing through the points $(2, 11)$ and $(5, 26)$ . Find the value of A and the value of B.

1
The equation is already a sum: the block $e^{2 y}$ equals A times the block $x^{2}$ , plus the constant B. So plot $e^{2 y}$ on the vertical axis and $x^{2}$ on the horizontal axis — no logarithms are needed here.
2
Define transformed variables $Y = e^{2 y}$ and $X = x^{2}$ . The relationship becomes $Y = A X + B$ , so the gradient is A and the intercept is B.
3
Find the gradient from the two given points: $A = \frac{26 - 11}{5 - 2} = \frac{15}{3} = 5$ .
4
Find B by substituting $(2, 11)$ into $Y = 5 X + B$ : $11 = 5 (2) + B$ , so $B = 11 - 10 = 1$ .
5
Check with the second point: $5 (5) + 1 = 26$ , which matches $(5, 26)$ . Therefore $A = 5$ and $B = 1$ .

📐 Worked Example

The variables x and y are related by $y = a x^{n}$ , where a and n are constants. Using natural logarithms, a plot of $ln y$ against $ln x$ is a straight line with gradient $0.5$ that passes through the point $(0, 1.6094)$ . Find the value of n and the value of a.

1
Take natural logs of both sides of $y = a x^{n}$ : $ln y = ln (a x^{n})$ .
2
Apply the product and power rules: $ln y = ln a + n ln x$ , which rearranges to $ln y = n ln x + ln a$ .
3
Match to $Y = m X + c$ with $Y = ln y$ and $X = ln x$ : the gradient is $m = n$ and the intercept is $c = ln a$ .
4
Read n from the gradient: $n = 0.5$ .
5
The line passes through $(0, 1.6094)$ , so the intercept is $ln a = 1.6094$ . Because natural logs were used, $a = e^{1.6094} = 5.00$ (3 s.f.).
6
So $n = 0.5$ and $a = 5$ , giving $y = 5 x^{0.5} = 5 x$ . Note the base rule: with $ln$ we recover $a = e^{c}$ , whereas with $l g$ we would use $a = 1 0^{c}$ .

✓ Quick check

For $y^{3} = A ln x + B$ , which quantities should you plot on the Y and X axes to obtain a straight line?
Reveal answer
$Y = y^3$ against $X = \ln x$ —
Correct! The $ln x$ term is already isolated and added to a constant, so plot $y^{3}$ against $ln x$ ; then gradient $= A$ and intercept $= B$ , with no further logs required.

Exam tip:

The $e$ and $ln$ forms appear regularly in 0606 papers. Spot them by an $e$ power on y or an $ln x$ term, then pick Y and X so no unknown is left trapped inside a logarithm or an exponent.

6. Working with Experimental Data Tables★★★★☆⏱ 5 min

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The most common linear-law exam question gives you a table of experimental values and asks you to (i) complete a row of transformed values, (ii) state which two quantities to plot to obtain a straight line, and (iii) use two points on that line to find the unknown constants. Work through the example below, which uses the exponential model $y = A b^{x}$ .

$x$	$y$
1	10
2	20
3	40
4	80
5	160

📐 Worked Example

The variables $x$ and $y$ in the table above are believed to satisfy $y = A b^{x}$ , where $A$ and $b$ are constants. (i) Complete a row of $l g y$ values, giving each to 2 decimal places. (ii) State which two quantities should be plotted to obtain a straight line. (iii) Using two points on the line, find the value of $A$ and the value of $b$ .

1
Part (i): Take $l g$ of each $y$ value: $l g 10 = 1.00$ , $l g 20 = 1.30$ , $l g 40 = 1.60$ , $l g 80 = 1.90$ , $l g 160 = 2.20$ (each to 2 d.p.).
2
Part (ii): Take $l g$ of both sides of $y = A b^{x}$ : $l g y = l g A + l g (b^{x}) = (l g b) x + l g A$ .
3
Compare with $Y = m X + c$ : plot $Y = l g y$ on the vertical axis against $X = x$ on the horizontal axis. The gradient is $l g b$ and the intercept is $l g A$ .
4
Part (iii): The transformed points $(x, l g y)$ lie on a straight line. Choose two points far apart, for example $(1, 1.00)$ and $(5, 2.20)$ .
5
$m = \frac{2.20 - 1.00}{5 - 1} = \frac{1.20}{4} = 0.30$
6
The gradient equals $l g b$ , so $l g b = 0.30$ and $b = 1 0^{0.30} \approx 2.0$ .
7
Find the intercept $c = l g A$ by substituting $(1, 1.00)$ into $l g y = 0.30 x + c$ : $1.00 = 0.30 (1) + c$ , so $c = 0.70$ .
8
Then $A = 1 0^{c} = 1 0^{0.70} \approx 5.0$ .
9
So $A \approx 5$ and $b \approx 2$ , giving $y \approx 5 (2)^{x}$ . Check with $x = 3$ : $5 (2)^{3} = 5 \times 8 = 40$ , which matches the table value.

Exam tip:

When reading two points off the line to find the gradient, choose points that are far apart and lie on clear grid intersections; this keeps rounding error small and protects your accuracy marks.

7. Drawing the Graph and Estimating Constants★★★★★⏱ 6 min

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In the previous section the two points were taken straight from the table. In the highest-mark version of these questions you are given graph paper: you must plot the transformed points yourself, draw a best-fit straight line by eye, then estimate the gradient and intercept by reading values off your drawn line. Because experimental points scatter slightly, these are estimates, so answers are usually given to 1-2 significant figures and a small range is accepted.

$x$	$y$
1	4.0
2	11.0
3	21.5
4	31.5
5	45.5

📐 Worked Example

The variables $x$ and $y$ in the table above are believed to satisfy $y = A x^{n}$ , where $A$ and $n$ are constants. By plotting $l g y$ against $l g x$ on graph paper, draw a straight-line graph and use it to estimate the value of $A$ and the value of $n$ .

1
Decide what to plot. Take $l g$ of both sides of $y = A x^{n}$ : $l g y = n l g x + l g A$ . Comparing with $Y = m X + c$ , plot $Y = l g y$ on the vertical axis against $X = l g x$ on the horizontal axis; the gradient is $n$ and the vertical-axis intercept is $l g A$ .
2
Build the transformed table, each value to 2 d.p. $l g x$ : $0.00, 0.30, 0.48, 0.60, 0.70$ . $l g y$ : $l g 4.0 = 0.60, l g 11.0 = 1.04, l g 21.5 = 1.33, l g 31.5 = 1.50, l g 45.5 = 1.66$ .
3
Plot the five points $(l g x, l g y)$ : $(0.00, 0.60), (0.30, 1.04), (0.48, 1.33), (0.60, 1.50), (0.70, 1.66)$ . They lie close to a straight line, so lay a ruler along them and draw a single best-fit line, keeping roughly equal numbers of points above and below it.
4
Estimate the gradient by reading two well-separated points that lie ON your drawn line (they need not be plotted data points). For example the line passes through about $(0.10, 0.75)$ and $(0.70, 1.65)$ .
5
$n = m \approx \frac{1.65 - 0.75}{0.70 - 0.10} = \frac{0.90}{0.60} = 1.5$
6
Estimate the intercept by reading where the drawn line crosses the vertical axis at $l g x = 0$ : here $c = l g A \approx 0.60$ .
7
Reverse the transformation to recover the constants: $A = 1 0^{c} = 1 0^{0.60} \approx 4.0$ and $n \approx 1.5$ .
8
State the result: $y \approx 4 x^{1.5}$ . Check against the table at $x = 4$ : $4 \times 4^{1.5} = 4 \times 8 = 32$ , close to the measured $31.5$ , so the estimate is sound.

✓ Quick check

To estimate the gradient from your best-fit line, which two points should you use?
Reveal answer
Two points that lie on the drawn line and are far apart —
Correct! Read the points off your ruled best-fit line (they need not be original data points) and choose them far apart so the gradient estimate is accurate.

Exam tip:

The highest-mark linear-law questions give you graph paper and ask you to plot the transformed points, draw one best-fit straight line, and read the constants off that line. Marks are awarded for accurate plotting, a sensible ruled line of best fit, and values estimated from the line — not for any table shortcut or regression formula.

8. Common Pitfalls

Wrong move:

Mixing up X variables for exponential vs power relationships

Why:

Power relationships use $l g x$ as X, exponential uses x as X, so mixing them gives incorrect gradient values

Correct move:

For $y = A x^{n}$ , $X = l g x$ ; for $y = A b^{x}$ , $X = x$ , always write out transformation steps to confirm

Wrong move:

Using $A = c$ directly instead of $A = 1 0^{c}$ when using log base 10

Why:

The intercept of the transformed graph is $l g A$ , not A itself

Correct move:

Always raise 10 (or e if using ln) to the power of the intercept to get the value of A

Wrong move:

Forgetting to apply logarithms to all terms on both sides

Why:

Missing the log on the constant A leads to an invalid linear form

Correct move:

Take the logarithm of every term on both sides when linearizing power/exponential relationships

Wrong move:

Using least squares regression to calculate gradient/intercept

Why:

Regression is out of scope for 0606, all questions provide a straight line or two points to calculate gradient from

Correct move:

Calculate gradient using two given points on the straight line, or use the stated gradient provided in the question

Wrong move:

Rearranging $y = A x^{n}$ to $y / x^{n} = A$ and claiming this is straight line form

Why:

This form has no X term with a gradient, so it does not match $Y = m X + c$

Correct move:

Use logarithms to linearize power and exponential relationships as demonstrated

9. Quick Reference Cheatsheet

Relationship Type	Original Equation	Transformed Y	Transformed X	Gradient m	Intercept c	Constant Calculation
Simple Polynomial	$y^{2} = A x^{3} + B$	$y^{2}$	$x^{3}$	A	B	$A = m$ , $B = c$
Power Relationship	$y = A x^{n}$	$l g y$	$l g x$	n	$l g A$	$n = m$ , $A = 1 0^{c}$
Exponential Relationship	$y = A b^{x}$	$l g y$	x	$l g b$	$l g A$	$b = 1 0^{m}$ , $A = 1 0^{c}$
Exponential in y	$e^{2 y} = A x^{2} + B$	$e^{2 y}$	$x^{2}$	A	B	$A = m$ , $B = c$
Power (natural log)	$y = a x^{n}$	$ln y$	$ln x$	n	$ln a$	$n = m$ , $a = e^{c}$
Log term in x	$y^{3} = A ln x + B$	$y^{3}$	$ln x$	A	B	$A = m$ , $B = c$

10. Frequently Asked

Do I use log base 10 or natural log for linearization?

For CIE 0606, log base 10 is standard unless specified otherwise. Both work as long as you are consistent: if using ln, calculate A as $e^{c}$ instead of $1 0^{c}$ .

Can I avoid using logs for all non-linear relationships?

Only for simple polynomial forms like $y^{2} = A x^{3} + B$ , which can be rearranged directly. Power and exponential relationships require logs to eliminate exponents on variables.

Going deeper

guideLogarithm Rules for 0606Prerequisite skill for linearizing power/exponential relationships

What's Next

Now that you have mastered linear law straight-line form, you can apply this skill to solve experimental data problems in both Paper 1 (non-calculator) and Paper 2 (calculator) of your CIE IGCSE Additional Mathematics exam. This skill is also foundational for future A-Level Mathematics topics like curve fitting and kinematics, if you choose to advance your studies. Next, practice solving past paper questions that combine linear law with logarithm rules and straight line graph calculations to build your speed and accuracy for the exam.