Reduction to Linear Form – OCR FSMQ Additional Maths

1 Why Reduce to a Straight Line?

EL4: convert equations of the form $y = ka^x$ and $y = kx^n$ to a linear form using logarithms. EL5: estimate the values of $k$ and $a$, or $k$ and $n$, from graphs.

The reason this is worth doing. Given a scatter of experimental points, it is easy to judge by eye whether they lie on a straight line, and easy to fit one with a ruler. Judging whether points fit a particular curve is far harder. So we transform the data until the relationship becomes a straight line, fit that line, and transform the answer back.

The target form

$Y = mX + c$, where $Y$ and $X$ are the transformed variables

Which transformation you need depends on the model. An exponential model needs logs of $y$ only; a power model needs logs of both. Identifying the model correctly is the first and most important step.

2 The Exponential Model $y = ka^x$

The transformation

$y = ka^x$
Take logs: $\log y = \log k + x\log a$
Compare with $Y = mX + c$: plot $\log y$ against $x$

Straight-line part	What it means
$Y$-axis	$\log y$
$X$-axis	$x$ — not transformed
Gradient $m$	$\log a$, so $a = 10^m$
Intercept $c$	$\log k$, so $k = 10^c$

Worked Example 1 — Finding $k$ and $a$ from a graph

Plotting $\log y$ against $x$ gives a straight line of gradient $0.3$ and intercept $0.7$. Find $k$ and $a$, and write down the model.

①Comparing with $\log y = \log k + x \log a$:

②$\log a = 0.3$, so $a = 10^{0.3} = 1.995$

③$\log k = 0.7$, so $k = 10^{0.7} = 5.012$

④$y \approx 5.01 \times 1.99^x$, or to sensible accuracy $y \approx 5 \times 2^x$.

Undo the logarithm with a power of $10$. The gradient and intercept are logarithms, so they must both be raised as powers of $10$ before they mean anything in the original model.

Worked Example 2 — Starting from data

The following data is believed to fit $y = ka^x$. Confirm this and find $k$ and $a$.

$x$	$1$	$2$	$3$	$4$
$y$	$6$	$18$	$54$	$162$
$\log y$	$0.778$	$1.255$	$1.732$	$2.210$

①The differences in $\log y$ are $0.477$, $0.477$, $0.478$ — constant, so $\log y$ against $x$ is a straight line ✓

②Gradient $= 0.477$, so $a = 10^{0.477} = 3.00$

③Intercept: extend back to $x = 0$. From $(1, 0.778)$ with gradient $0.477$: $c = 0.778 - 0.477 = 0.301$

④$k = 10^{0.301} = 2.00$

⑤$y = 2 \times 3^x$

⑥Check: $x=1 \to 6$ ✓, $x=4 \to 2 \times 81 = 162$ ✓

Constant differences in $\log y$ confirm the model. This is the neatest way to verify an exponential relationship, and it also gives the gradient directly.

3 The Power Model $y = kx^n$

The transformation

$y = kx^n$
Take logs: $\log y = \log k + n\log x$
Compare with $Y = mX + c$: plot $\log y$ against $\log x$

Straight-line part	What it means
$Y$-axis	$\log y$
$X$-axis	$\log x$ — also transformed
Gradient $m$	$n$ directly — no undoing needed
Intercept $c$	$\log k$, so $k = 10^c$

The one distinction to hold on to

Exponential $ka^x$: plot $\log y$ against $x$; the gradient is $\log a$
Power $kx^n$: plot $\log y$ against $\log x$; the gradient is $n$

Worked Example 3 — A power model from a graph

Plotting $\log y$ against $\log x$ gives a straight line of gradient $1.5$ and intercept $0.6$. Find $n$ and $k$.

①The gradient is $n$ directly, so $n = 1.5$.

②$\log k = 0.6$, so $k = 10^{0.6} = 3.98$

③$y \approx 3.98\, x^{1.5}$

Do not raise the gradient to a power of $10$ here. In a power model the gradient is already the exponent $n$. Only the intercept needs undoing. Mixing these two cases up is the commonest error in this topic.

Worked Example 4 — From data, with a physical interpretation

A pendulum's period $T$ seconds is measured against its length $L$ metres. Test the model $T = kL^n$ and find $k$ and $n$.

$L$	$0.25$	$1$	$4$
$T$	$1.00$	$2.00$	$4.00$
$\log L$	$-0.602$	$0$	$0.602$
$\log T$	$0$	$0.301$	$0.602$

①Gradient: $\dfrac{0.602 - 0}{0.602 - (-0.602)} = \dfrac{0.602}{1.204} = 0.5$

②So $n = 0.5$, i.e. $T \propto \sqrt{L}$.

③Intercept at $\log L = 0$: $\log T = 0.301$, so $k = 10^{0.301} = 2.00$

④$T = 2\sqrt{L}$

⑤Check: $L = 4$ gives $T = 2 \times 2 = 4$ ✓

A gradient of $\tfrac12$ means a square-root law. This is exactly the physics result $T = 2\pi\sqrt{L/g}$, whose constant $\tfrac{2\pi}{\sqrt{9.8}} = 2.006$ — a striking confirmation that the log–log method recovers real physical laws from raw measurements.

4 Deciding Which Model Fits

Tabulate $x$, $y$, $\log x$ and $\log y$.
Check whether $\log y$ against $x$ looks straight — if so, the model is exponential.
Check whether $\log y$ against $\log x$ looks straight — if so, the model is a power law.
Fit the straight line, read the gradient and intercept, and convert back.
Test the model against a data point you did not use.

Worked Example 5 — Distinguishing the two

For each set of data, decide whether the relationship is exponential or a power law.

①Set A: $(1,3), (2,9), (3,27), (4,81)$. Each $y$ is $3$ times the last, at equal $x$ steps.

②Equal additive steps in $x$ giving equal multiplicative steps in $y$ is the signature of an exponential: $y = 3^x$.

③Set B: $(1,2), (2,8), (3,18), (4,32)$. Here $y \div x^2 = 2$ every time.

④That constant ratio to a power of $x$ is the signature of a power law: $y = 2x^2$.

⑤Note that doubling $x$ in Set B multiplies $y$ by $4$ — a fixed factor for a fixed ratio of $x$, rather than a fixed difference.

The quick test: if equal steps in $x$ multiply $y$ by a constant, it is exponential. If equal ratios of $x$ multiply $y$ by a constant, it is a power law.

5 Quick Reference

Target

$Y = mX + c$.

$y = ka^x$

$\log y = \log k + x\log a$.

Plot

$\log y$ against $x$.

Then

$a = 10^{\text{gradient}}$, $k = 10^{\text{intercept}}$.

$y = kx^n$

$\log y = \log k + n\log x$.

Plot

$\log y$ against $\log x$.

Then

$n = $ gradient directly; $k = 10^{\text{intercept}}$.

The key difference

Power: gradient needs no undoing.

Verifying

Constant differences in $\log y$ $\Rightarrow$ exponential.

Always

Test the model on an unused point.

6 Practice Questions

Question 1

What should be plotted against what to linearise $y = ka^x$?

▶ Show solution

Plot $\log y$ (vertical) against $x$ (horizontal).

Only $y$ is transformed.

Question 2

What should be plotted against what to linearise $y = kx^n$?

▶ Show solution

Plot $\log y$ against $\log x$ — both variables are transformed.

Question 3

Plotting $\log y$ against $x$ gives gradient $0.5$ and intercept $1$. Find $k$ and $a$.

▶ Show solution

$a = 10^{0.5} = 3.16$ (3 s.f.)

$k = 10^1 = 10$

$y = 10 \times 3.16^x$

Question 4

Plotting $\log y$ against $\log x$ gives gradient $2$ and intercept $0.48$. Find $k$ and $n$.

▶ Show solution

$n = 2$ (the gradient directly).

$k = 10^{0.48} = 3.02$ (3 s.f.)

$y \approx 3x^2$

Question 5

Write $y = 5 \times 4^x$ in linear form.

▶ Show solution

$\log y = \log 5 + x\log 4$

$= 0.699 + 0.602x$

A straight line of gradient $0.602$ and intercept $0.699$ when $\log y$ is plotted against $x$.

Question 6

Write $y = 3x^4$ in linear form.

▶ Show solution

$\log y = \log 3 + 4\log x$

$= 0.477 + 4\log x$

Gradient $4$, intercept $0.477$, plotting $\log y$ against $\log x$.

Question 7

A log–log plot has gradient $0.5$. What does this tell you about the relationship?

▶ Show solution

$n = 0.5$, so $y = kx^{0.5} = k\sqrt{x}$.

The relationship is a square-root law: quadrupling $x$ doubles $y$.

Question 8

Data gives $\log y$ values of $0.5$, $0.8$, $1.1$, $1.4$ at $x = 1, 2, 3, 4$. Find the model.

▶ Show solution

Differences in $\log y$ are all $0.3$, so plotting against $x$ gives a straight line of gradient $0.3$ — an exponential model.

$a = 10^{0.3} = 2.00$

Intercept: from $(1, 0.5)$ back to $x=0$ gives $c = 0.5 - 0.3 = 0.2$, so $k = 10^{0.2} = 1.58$.

$y \approx 1.58 \times 2^x$

Question 9

Explain why plotting $\log y$ against $x$ for a power-law relationship would not give a straight line.

▶ Show solution

For $y = kx^n$, taking logarithms gives $\log y = \log k + n\log x$.

The right-hand side involves $\log x$, not $x$. Plotting against $x$ would therefore graph $\log k + n\log x$ against $x$, which is still a curve (a logarithmic one).

Only plotting against $\log x$ makes the right-hand side linear in the horizontal variable.

Question 10

The braking distance $d$ metres of a car at speed $v$ mph is recorded:

$v$	$20$	$40$	$60$	$80$
$d$	$6.0$	$24.0$	$54.0$	$96.0$

(a) Explain why a power model $d = kv^n$ is more likely than an exponential one. (b) Complete a table of $\log v$ and $\log d$, and find $n$ and $k$. (c) State the model and check it against a data point. (d) Use it to predict the braking distance at $70$ mph, and comment on the reliability of that prediction.

▶ Show solution

(a) Doubling $v$ from $20$ to $40$ multiplies $d$ by $4$; doubling again from $40$ to $80$ multiplies $d$ by $4$ once more ($24 \to 96$).

A constant factor for a constant ratio of $v$ is the signature of a power law. An exponential would need a constant factor for a constant difference in $v$, and here the step from $20$ to $40$ ($\times 4$) differs from $40$ to $60$ ($\times 2.25$).

(b)

$v$	$20$	$40$	$60$	$80$
$\log v$	$1.301$	$1.602$	$1.778$	$1.903$
$\log d$	$0.778$	$1.380$	$1.732$	$1.982$

Gradient, using the first and last points:

$n = \dfrac{1.982 - 0.778}{1.903 - 1.301} = \dfrac{1.204}{0.602} = 2.00$

Intercept: from $(1.301, 0.778)$ with gradient $2$:

$c = 0.778 - 2(1.301) = 0.778 - 2.602 = -1.824$

$k = 10^{-1.824} = 0.0150$

(c) $d = 0.015\,v^2$

Check at $v = 60$ (a point not used for the gradient): $0.015 \times 3600 = 54.0$ ✓ — exactly the recorded value.

(d) At $v = 70$: $d = 0.015 \times 4900 = 73.5$ m.

Reliability: this is an interpolation — $70$ mph lies between the data values $60$ and $80$ — and the model fitted every recorded point exactly. So the prediction is well founded.

Extrapolating far beyond the data would be much less safe: at $150$ mph the model gives $337$ m, but tyre and brake behaviour changes at high speed, so the simple $v^2$ law may no longer hold. Predictions inside the range of the data are always more trustworthy than those outside it.

Reduction to Linear Form (EL4–EL5) · OCR FSMQ Additional Maths · Created with MathJax