# Linear Law — Straight-Line Form

> CIE IGCSE Additional Mathematics · 0606 2025-2027
> Source: https://www.owlsprep.com/study/cie-0606-u7-linear-law-straight-line-form/

This guide teaches you to convert non-linear mathematical relationships into the standard straight-line form $Y = mX + c$ using algebraic rearrangement and logarithms, per CIE IGCSE Additional Mathematics 0606 syllabus requirements.

**Prerequisites:** Knowledge of straight line graphs (gradient, intercept, $Y=mX+c$ form); Proficiency with logarithm rules (power, product)

## Learning objectives

- Transform non-linear relationships y=Axⁿ, y=Abˣ, y²=Ax³+B into Y=mX+c form
- Identify Y, X, gradient m and intercept c for each transformed relationship
- Calculate constants A, n, b, B from the gradient and intercept of transformed graphs
- Apply linear law to solve structured CIE 0606 exam questions correctly
- Complete transformed rows in experimental data tables and find constants from two points on the fitted line

## Transforming Polynomial Relationships

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 $y^2 = Ax^3 + B$.

**Straight-Line Form Transformation** — The process of rearranging a non-linear relationship to match $Y = mX + c$, 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 $y^2 = 5x^3 + 2$ into $Y = mX + c$ 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: $Y = 5X + 2$.
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.

## Linearizing Power Relationships ($y = Ax^n$)

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.

> **tip**
>
> If you use natural log (ln) instead of log₁₀, you will calculate $A = e^c$ instead of $10^c$. Always use the same log base across all your working.

**Worked example:** Transform $y = Ax^n$ 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: $\lg y = \lg (Ax^n)$.
2. Apply logarithm product rule: $\lg y = \lg A + \lg x^n$.
3. Apply logarithm power rule: $\lg y = n \lg x + \lg A$.
4. Match to $Y = mX + c$: $Y = \lg y$, $X = \lg x$, gradient $m = n$, intercept $c = \lg A$.
5. Calculate constants: n equals the gradient of the graph, $A = 10^c$ where c is the Y-intercept.

## Linearizing Exponential Relationships ($y = Ab^x$)

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

1. Take log₁₀ of both sides: $\lg y = \lg (Ab^x)$.
2. Apply product rule: $\lg y = \lg A + \lg b^x$.
3. Apply power rule: $\lg y = (\lg b) x + \lg A$.
4. Match to $Y = mX + c$: $Y = \lg y$, $X = x$, gradient $m = \lg b$, intercept $c = \lg A$.
5. Calculate constants: $b = 10^m$, $A = 10^c$.

**Check your understanding**

1. For the relationship $y = 3(2)^x$, what is the gradient of the transformed graph of $\lg y$ against x?

   *Why:* Correct! The gradient equals $\lg b$ where b is the base of the exponential, here b=2.

## Solving Linear Law Exam Problems

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 $\lg y$ against $\lg x$ gives a straight line with gradient 2 and Y-intercept 0.69897. Find the values of A and n for the relationship $y = Ax^n$.

1. Recall the transformed form for power relationships: $\lg y = n \lg x + \lg A$.
2. Match gradient to n: $n = 2$.
3. Match intercept to $\lg A$: $\lg A = 0.69897$.
4. Calculate A: $A = 10^{0.69897} = 5$.
5. Final relationship: $y = 5x^2$.

**Exam command terms**

- **Transform into straight line form** — Rearrange the given non-linear equation to match $Y = mX + c$, clearly stating your definitions of Y and X. *(For $y = 7x^3$, write $Y = \lg y$, $X = \lg x$, so $Y = 3X + \lg7$.)*

- **Find the constants** — Use the gradient and intercept of the transformed graph to calculate the unknown values in the original non-linear relationship.

*Calculator:* allowed

## Linearizing $e$ and $\ln$ Relationships

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 $e^{2y} = Ax^2 + B$ and $y^3 = A\ln x + B$. Handle these exactly like the earlier cases: choose transformed variables Y and X so the equation becomes $Y = mX + c$, then read the unknown constants from the gradient and intercept.

> **tip**
>
> When a whole $e^{ky}$ block, or an $\ln x$ term, is simply added to a constant, no logarithms are needed — plot that block directly. Only take a logarithm when an unknown is trapped inside an exponent or a power.

**Worked example:** The variables x and y satisfy $e^{2y} = Ax^2 + B$, where A and B are constants. When $e^{2y}$ 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 $e^{2y}$ equals A times the block $x^2$, plus the constant B. So plot $e^{2y}$ on the vertical axis and $x^2$ on the horizontal axis — no logarithms are needed here.
2. Define transformed variables $Y = e^{2y}$ and $X = x^2$. The relationship becomes $Y = AX + B$, 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 $Y = 5X + B$: $11 = 5(2) + B$, 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 $y = ax^n$, 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 $y = ax^n$: $\ln y = \ln(ax^n)$.
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 $Y = mX + c$ 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 $y = 5x^{0.5} = 5\sqrt{x}$. Note the base rule: with $\ln$ we recover $a = e^{c}$, whereas with $\lg$ we would use $a = 10^{c}$.

**Check your understanding**

1. For $y^3 = A\ln x + B$, which quantities should you plot on the Y and X axes to obtain a straight line?

   *Why:* 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.

*Calculator:* allowed

## Working with Experimental Data Tables

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 $y = Ab^x$.

| $x$ | $y$ |
| --- | --- |
| 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 $y = Ab^x$, where $A$ and $b$ are constants. (i) Complete a row of $\lg y$ 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 $\lg$ of each $y$ value: $\lg 10 = 1.00$, $\lg 20 = 1.30$, $\lg 40 = 1.60$, $\lg 80 = 1.90$, $\lg 160 = 2.20$ (each to 2 d.p.).
2. Part (ii): Take $\lg$ of both sides of $y = Ab^x$: $\lg y = \lg A + \lg(b^x) = (\lg b)x + \lg A$.
3. Compare with $Y = mX + c$: plot $Y = \lg y$ on the vertical axis against $X = x$ on the horizontal axis. The gradient is $\lg b$ and the intercept is $\lg A$.
4. Part (iii): The transformed points $(x, \lg y)$ 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 $\lg b$, so $\lg b = 0.30$ and $b = 10^{0.30} \approx 2.0$.
7. Find the intercept $c = \lg A$ by substituting $(1, 1.00)$ into $\lg y = 0.30x + c$: $1.00 = 0.30(1) + c$, so $c = 0.70$.
8. Then $A = 10^{c} = 10^{0.70} \approx 5.0$.
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.

> **tip**
>
> For a power model $y = Ax^n$ the same routine applies, but you must also complete a row of $\lg x$ and plot $\lg y$ against $\lg x$. For $y = Ab^x$ only $\lg y$ is needed, because $X = x$ is used unchanged.

> **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.

*Calculator:* allowed

## Drawing the Graph and Estimating Constants

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 $y = Ax^n$, where $A$ and $n$ are constants. By plotting $\lg y$ against $\lg x$ 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 $\lg$ of both sides of $y = Ax^n$: $\lg y = n\lg x + \lg A$. Comparing with $Y = mX + c$, plot $Y = \lg y$ on the vertical axis against $X = \lg x$ on the horizontal axis; the gradient is $n$ and the vertical-axis intercept is $\lg A$.
2. Build the transformed table, each value to 2 d.p. $\lg x$: $0.00, 0.30, 0.48, 0.60, 0.70$. $\lg y$: $\lg 4.0 = 0.60, \lg 11.0 = 1.04, \lg 21.5 = 1.33, \lg 31.5 = 1.50, \lg 45.5 = 1.66$.
3. Plot the five points $(\lg x, \lg y)$: $(0.00, 0.60), (0.30, 1.04), (0.48, 1.33), (0.60, 1.50), (0.70, 1.66)$. 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 $\lg x = 0$: here $c = \lg A \approx 0.60$.
7. Reverse the transformation to recover the constants: $A = 10^{c} = 10^{0.60} \approx 4.0$ and $n \approx 1.5$.
8. State the result: $y \approx 4x^{1.5}$. 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.

> **tip**
>
> Only read the intercept directly where the horizontal-axis value is truly $0$. If your $X$-axis does not start at $0$ (a broken scale), do not read the crossing at the left edge of the grid — instead find the gradient first, then substitute one point from the line into $Y = mX + c$ to solve for $c$.

**Check your understanding**

1. To estimate the gradient from your best-fit line, which two points should you use?

   *Why:* 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.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Mixing up X variables for exponential vs power relationships
  - Why it fails: Power relationships use $\lg x$ as X, exponential uses x as X, so mixing them gives incorrect gradient values
  - Correct: For $y=Ax^n$, $X=\lg x$; for $y=Ab^x$, $X=x$, always write out transformation steps to confirm
- **Wrong:** Using $A = c$ directly instead of $A = 10^c$ when using log base 10
  - Why it fails: The intercept of the transformed graph is $\lg A$, not A itself
  - Correct: Always raise 10 (or e if using ln) to the power of the intercept to get the value of A
- **Wrong:** Forgetting to apply logarithms to all terms on both sides
  - Why it fails: Missing the log on the constant A leads to an invalid linear form
  - Correct: Take the logarithm of every term on both sides when linearizing power/exponential relationships
- **Wrong:** Using least squares regression to calculate gradient/intercept
  - Why it fails: Regression is out of scope for 0606, all questions provide a straight line or two points to calculate gradient from
  - Correct: Calculate gradient using two given points on the straight line, or use the stated gradient provided in the question
- **Wrong:** Rearranging $y=Ax^n$ to $y/x^n = A$ and claiming this is straight line form
  - Why it fails: This form has no X term with a gradient, so it does not match $Y=mX+c$
  - Correct: Use logarithms to linearize power and exponential relationships as demonstrated

## Cheatsheet

| Relationship Type | Original Equation | Transformed Y | Transformed X | Gradient m | Intercept c | Constant Calculation |
| --- | --- | --- | --- | --- | --- | --- |
| Simple Polynomial | $y^2 = Ax^3 + B$ | $y^2$ | $x^3$ | A | B | $A=m$, $B=c$ |
| Power Relationship | $y = Ax^n$ | $\lg y$ | $\lg x$ | n | $\lg A$ | $n=m$, $A=10^c$ |
| Exponential Relationship | $y = Ab^x$ | $\lg y$ | x | $\lg b$ | $\lg A$ | $b=10^m$, $A=10^c$ |
| Exponential in y | $e^{2y} = Ax^2 + B$ | $e^{2y}$ | $x^2$ | A | B | $A=m$, $B=c$ |
| Power (natural log) | $y = ax^n$ | $\ln y$ | $\ln x$ | n | $\ln a$ | $n=m$, $a=e^c$ |
| Log term in x | $y^3 = A\ln x + B$ | $y^3$ | $\ln x$ | A | B | $A=m$, $B=c$ |

## 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.

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