Modelling Curves

A2 · P3 · 3 min

Many relationships between two variables are not linear, but become linear after taking logarithms. Plotting the transformed data as a straight line lets you read the unknown constants from the gradient and intercept. The syllabus requires two models: a power law and an exponential law.

The two models

Key result

Power law $y = kx^n$ :

\ln y = \ln k + n\ln x

Plot $\ln y$ against $\ln x$ : a straight line with gradient $n$ and intercept $\ln k$ .

Exponential law $y = k\,a^x$ :

\ln y = \ln k + x\ln a

Plot $\ln y$ against $x$ : a straight line with gradient $\ln a$ and intercept $\ln k$ .

Any base of logarithm works, as long as you use the same base throughout; questions may use $\log_{10}$ or $\ln$ .

Method

Take logs of both sides and use the laws to write the equation in the form $Y = mX + c$ .
Identify what must be plotted ( $X$ and $Y$ ) and what the gradient and intercept represent.
From the graph or from two points, find the gradient and intercept.
Convert back: $k = e^{c}$ (or $10^c$ ), and $a = e^{m}$ for the exponential model, $n = m$ for the power model.

Power law from a line

Variables $x$ and $y$ satisfy $y = kx^n$ . The graph of $\ln y$ against $\ln x$ is a straight line through $(1, 3.2)$ and $(3, 4.4)$ . Find $k$ and $n$ .

Solution

Gradient $n = \dfrac{4.4 - 3.2}{3 - 1} = 0.6$ .

Intercept: $\ln y = 0.6\ln x + c$ ; using $(1, 3.2)$ : $c = 3.2 - 0.6 = 2.6$ . So $\ln k = 2.6$ , $k = e^{2.6} \approx 13.5$ .

$y \approx 13.5x^{0.6}$ .

Exponential law from data

The table gives values of $y$ for values of $x$ :

$x$	1	2	3	4
$y$	6.1	9.2	13.7	20.6

It is thought that $y = k a^x$ . Show how a linear graph can be obtained, and estimate $k$ and $a$ .

Solution

$\ln y = \ln k + x\ln a$ , so plot $\ln y$ against $x$ : values $1.81, 2.22, 2.62, 3.03$ , which increase by about $0.405$ each step, confirming a straight line.

Gradient $\ln a \approx 0.405 \Rightarrow a \approx e^{0.405} = 1.5$ . Intercept: $\ln k \approx 1.81 - 0.405 = 1.405 \Rightarrow k \approx 4.1$ .

$y \approx 4.1 \times 1.5^x$ .

Reading the intercept when the line is given

The variables satisfy $\log_{10} y = 0.5\log_{10} x + 1.3$ . Express $y$ in terms of $x$ .

Solution

$\log_{10} y = \log_{10} x^{0.5} + 1.3 \Rightarrow y = 10^{1.3}\sqrt{x} \approx 20.0\sqrt{x}$ .

Which graph?

For each model, state the graph that gives a straight line and what the gradient represents: (a) $y = Ae^{bx}$ ; (b) $y = Ax^b$ ; (c) $y^2 = Ax^b$ .

Solution

(a) $\ln y$ against $x$ ; gradient $b$ . (b) $\ln y$ against $\ln x$ ; gradient $b$ . (c) $2\ln y = \ln A + b\ln x$ , so $\ln y$ against $\ln x$ has gradient $\tfrac{b}{2}$ .

Watch out

The intercept of the transformed graph is $\ln k$ , not $k$ . Forgetting to exponentiate is the most common error in this topic. Similarly, for an exponential model the gradient is $\ln a$ , so $a = e^{\text{gradient}}$ .

Exam tip

Read gradients from two points on the drawn line, not from the raw data points, and quote them to about 2 significant figures. The mark scheme allows a tolerance because the line is fitted by eye.

Practice

Question

$y = kx^n$ . The line of $\ln y$ against $\ln x$ has gradient $-2$ and intercept $1.5$ . Find $k$ and $n$ .
$y = ka^x$ . The line of $\ln y$ against $x$ passes through $(0, 0.7)$ and $(5, 2.2)$ . Find $k$ and $a$ .
$y = 3 \times 2^{x}$ . State the gradient and intercept of the graph of $\log_{10} y$ against $x$ .
$p = kq^n$ with $p = 12$ when $q = 2$ and $p = 48$ when $q = 8$ . Find $n$ and $k$ .

Answers

$n = -2$ , $k = e^{1.5} \approx 4.48$ .
Gradient $0.3$ , so $a = e^{0.3} \approx 1.35$ ; $k = e^{0.7} \approx 2.01$ .
Gradient $\log_{10} 2 \approx 0.301$ ; intercept $\log_{10} 3 \approx 0.477$ .
$\dfrac{48}{12} = 4^n \Rightarrow n = 1$ ; $k = 6$ .

The two models

Practice

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