Transform a non-linear relationship into the form Y equals mX plus c and use the gradient and intercept of the straight-line graph to find unknown constants

What this dot point is asking

SEAB wants you to take a non-linear law connecting two variables, such as $y = ax^n$ or $y = Ab^x$, and rewrite it so that plotting suitable transformed quantities gives a straight line $Y = mX + c$. From the gradient and intercept of that line, read off the unknown constants. This is how experimental data is used to confirm a law and measure its parameters.

The answer

Why we linearise

A straight line is the easiest graph to draw and read. If a relationship can be cast as $Y = mX + c$ for some functions $Y$ and $X$ of the original variables, then plotting $Y$ against $X$ gives a line whose gradient $m$ and intercept $c$ carry the unknown constants.

The power law $y = ax^n$

Take logarithms (base $10$ or $e$) of both sides:

Plotting $\lg y$ against $\lg x$ gives a line of gradient $n$ and intercept $\lg a$.

The exponential law $y = Ab^x$

Take logarithms of both sides:

Plotting $\lg y$ against $x$ (note: $x$ itself, not $\lg x$) gives a line of gradient $\lg b$ and intercept $\lg A$.

Recovering the constants

Match the linear form to $Y = mX + c$, identify what $Y$ and $X$ stand for, then solve: an intercept of $\lg a$ gives $a = 10^{\,\text{intercept}}$, and a gradient equal to $\lg b$ gives $b = 10^{\,\text{gradient}}$.

Choosing what to plot for an unfamiliar law

Not every relationship is a clean power or exponential law, so the general skill is to rearrange any equation into the shape $Y = mX + c$ and read off what $Y$ and $X$ must be. For $y = ax^2 + b$, no logarithms are needed: plotting $y$ against $x^2$ is already linear, with gradient $a$ and intercept $b$. For $\tfrac{1}{y} = ax + b$, plot $\tfrac{1}{y}$ against $x$. The method is to isolate the part containing the unknown constants as a linear combination of two computable quantities, then those two quantities become your axes. Recognising that linearising is just "force it into $Y = mX + c$" extends the technique well beyond the two standard laws.

Reading constants from a best-fit line through data

With real experimental points, the plotted data will not lie perfectly on a line, so you draw a line of best fit and take its gradient and intercept from two well-separated points on that line, not from raw data points. Using points far apart on the fitted line reduces the effect of reading errors on the gradient. Then convert as usual, for instance $a = 10^{\text{intercept}}$. Emphasising the best-fit line, rather than any single data point, is what makes the constants reliable, and it is the practical reason the linear law is so useful for analysing measurements.

Examples in context

Example 1. Verifying a physical law. To test whether a pendulum's period $T$ obeys $T = kL^p$, experimenters plot $\lg T$ against $\lg L$; a straight line confirms the power law, and the gradient gives the exponent $p$ (close to $0.5$), validating the theory.

Example 2. Modelling bacterial growth. Counting bacteria over time and plotting $\lg N$ against $t$ gives a straight line if growth is exponential, $N = N_0 b^t$; the gradient measures the growth rate through $\lg b$.

Try this

Q1. Write $y = ax^n$ in straight-line form and state what is plotted on each axis. [2 marks]

• Cue. $\lg y = n\lg x + \lg a$; plot $\lg y$ (vertical) against $\lg x$ (horizontal).

Q2. A plot of $\lg y$ against $x$ for $y = Ab^x$ has gradient $0.5$. Find $b$. [2 marks]

• Cue. $\lg b = 0.5$, so $b = 10^{0.5} = \sqrt{10} \approx 3.16$.

Q3. For $y = ax^n$, the $\lg y$-against-$\lg x$ line has intercept $2$. Find $a$. [2 marks]

• Cue. $\lg a = 2$, so $a = 100$.