Linear Law: The Intercept Is lg a, Not a

The question

The variables $x$ and $y$ are related by the equation

y = a x^n

where $a$ and $n$ are constants. When $\lg y$ is plotted against $\lg x$ , a straight line is obtained. The line passes through the points $(0.4,\ 1.20)$ and $(1.2,\ 2.40)$ , as shown in the diagram. ( $\lg$ means $\log_{10}$ .)

Straight-line graph of lg y against lg x. The line passes through the points (0.4, 1.20) and (1.2, 2.40) and crosses the lg y axis above the origin. — The graph of lg y against lg x is a straight line through (0.4, 1.20) and (1.2, 2.40).

(a) Find the value of $n$ and the value of $a$ , giving $a$ to 3 significant figures. [4 marks]

(b) Find the value of $y$ when $x = 5$ . [2 marks]

The common wrong answer

Most students get the gradient right:

n = \frac{2.40 - 1.20}{1.2 - 0.4} = \frac{1.20}{0.8} = 1.5

They also find the intercept correctly. Using the point $(0.4,\ 1.20)$ in $Y = mX + c$ :

1.20 = 1.5(0.4) + c \quad \Rightarrow \quad c = 0.6

Then comes the slip. They see that the intercept “goes with” $a$ and write $a = 0.6$ , so the equation becomes $y = 0.6x^{1.5}$ . Everything after that is wrong:

\begin{aligned} \text{(b)}\quad y &= 0.6 \times 5^{1.5} = 0.6 \times 11.18 = 6.71 \\ \text{(c)}\quad 100 &= 0.6x^{1.5} \ \Rightarrow \ x^{1.5} = 166.7 \ \Rightarrow \ x = 30.3 \end{aligned}

The mistake is forgetting what is on the vertical axis. The graph is of $\lg y$ , not $y$ , so its intercept is $\lg a$ , not $a$ . The Cambridge 0606 examiner report for June 2024 described the same problem on a log graph question: more commonly than a correct equation, candidates wrote " $y =$ " where the equation of the line needed " $\ln y =$ ". Once the log is dropped from one side, the constant comes out wrong.

You can catch it with a quick check. If $a$ really were 0.6, the intercept would be $\lg 0.6 = -0.222$ . That is below the origin, but the line crosses the axis above it.

The correct working

(a) Take logs of both sides of $y = ax^n$ and use the log laws:

\lg y = \lg a + n \lg x

Rearrange it into the form $Y = mX + c$ :

\underbrace{\lg y}_{Y} = n\,\underbrace{\lg x}_{X} + \underbrace{\lg a}_{c}

So the gradient is $n$ and the intercept on the $\lg y$ axis is $\lg a$ .

Gradient:

n = \frac{2.40 - 1.20}{1.2 - 0.4} = \frac{1.20}{0.8} = 1.5

Intercept, using the point $(0.4,\ 1.20)$ :

\begin{aligned} 1.20 &= 1.5(0.4) + \lg a \\ \lg a &= 1.20 - 0.6 = 0.6 \end{aligned}

Undo the log to get $a$ :

a = 10^{0.6} = 3.981\ldots

So $n = 1.5$ and $a = 3.98$ (3 s.f.), which gives $y = 3.98x^{1.5}$ .

Check with the other point: $1.5(1.2) + 0.6 = 2.40$ . It lies on the line.

(b) Use the straight-line equation with $x = 5$ , so $\lg x = \lg 5 = 0.6990$ :

\begin{aligned} \lg y &= 1.5(0.6990) + 0.6 = 1.6485 \\ y &= 10^{1.6485} = 44.5 \end{aligned}

So $y = 44.5$ (3 s.f.). You get the same from $y = 3.98 \times 5^{1.5} = 3.98 \times 11.18 = 44.5$ .

(c) When $y = 100$ , $\lg y = \lg 100 = 2$ :

\begin{aligned} 2 &= 1.5 \lg x + 0.6 \\ \lg x &= \frac{1.4}{1.5} = 0.9333 \\ x &= 10^{0.9333} = 8.58 \end{aligned}

So $x = 8.58$ (3 s.f.). Check: $3.981 \times 8.577^{1.5} = 100$ .

Why it works

Taking logs turns a curve into a straight line, but it changes what the axes measure. Every point on this graph is $(\lg x,\ \lg y)$ , not $(x,\ y)$ . The point $(1.2,\ 2.40)$ , for example, means $x = 10^{1.2} \approx 15.8$ and $y = 10^{2.40} \approx 251$ .

The intercept is the value of $\lg y$ when $\lg x = 0$ , that is, when $x = 1$ . Put $x = 1$ into $y = ax^n$ and you get $y = a$ , so the intercept is $\lg a$ . To get $a$ itself you have to undo the log: $a = 10^{\text{intercept}}$ .

The gradient is different. The power $n$ is the coefficient of $\lg x$ , so it comes straight off the graph with no log to undo. That is why students who get $n$ right can still get $a$ wrong.

In parts (b) and (c) it is often safest to stay with the straight-line equation, $\lg y = 1.5\lg x + 0.6$ . It uses the exact intercept, so you avoid carrying the rounded $a = 3.98$ .

How to spot it next time

Before reading anything off the graph, write the log form with each part labelled: $\lg y = n\lg x + \lg a$ . Whatever sits in the $c$ position is what the intercept equals.
If the vertical axis is $\lg y$ , the intercept is a log. Raise 10 to that power to get the constant. If the axis is $\ln y$ , use $e$ instead.
Check your final equation with a point from the graph. Convert it back first: the point $(0.4,\ 1.20)$ means $x = 10^{0.4}$ and $y = 10^{1.20}$ .