AS & A Level · Practice questions · Reducing relationships to linear form

Logs flatten curves.

Papers 2 and 3. Six original Cambridge-style questions on reducing \(y = kx^n\) and \(y = ka^x\) to linear form, ending with a growth model fitted with base-10 logs.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Take logs, match \(Y = mX + c\), exponentiate.

\(y = kx^n\): \(\ln y\) against \(\ln x\); gradient \(n\), intercept \(\ln k\).
\(y = ka^x\): \(\ln y\) against \(x\); gradient \(\ln a\), intercept \(\ln k\).
Gradient from two points, then intercept from \(Y = mX + c\) at one point.
Undo the logs: \(k = e^{c}\), \(a = e^{m}\) (or \(10^{(\dots)}\) for lg).
Answer the question: write the model, then substitute if asked.

Recall

[4 marks]

For each relationship, state which variables should be plotted to give a straight line, and give the gradient and intercept: (a) \(y = 5x^3\) (b) \(y = 2 \times 7^x\).

(a) \(\ln y = 3\ln x + \ln 5\): plot \(\ln y\) against \(\ln x\); gradient \(3\), intercept \(\ln 5\). ✓
(b) \(\ln y = (\ln 7)x + \ln 2\): plot \(\ln y\) against \(x\); gradient \(\ln 7\), intercept \(\ln 2\). ✓

Calculation

[4 marks]

The variables \(x\) and \(y\) satisfy \(y = kx^n\). The graph of \(\ln y\) against \(\ln x\) is a straight line through \((1.0, 3.2)\) and \((2.5, 2.3)\). Find \(k\) and \(n\).

Gradient: \(\dfrac{2.3 - 3.2}{2.5 - 1.0} = \dfrac{-0.9}{1.5} = -0.6\), so \(n = -0.6\). ✓
Intercept: \(3.2 = -0.6(1.0) + c\), so \(c = 3.8 = \ln k\). ✓
\(k = e^{3.8} = 44.7\) (3 s.f.), so \(y = 44.7x^{-0.6}\). ✓

Calculation

[4 marks]

The variables satisfy \(y = Ae^{kx}\). The graph of \(\ln y\) against \(x\) passes through \((0, 2.4)\) and has gradient \(-0.3\).

(a) Find \(A\) and \(k\). (b) Find \(y\) when \(x = 4\). (c) Find \(x\) when \(y = 5\).

(a) \(k = -0.3\) and \(\ln A = 2.4\), so \(A = e^{2.4} = 11.0\) (3 s.f.). ✓
(b) \(\ln y = 2.4 - 0.3(4) = 1.2\), so \(y = e^{1.2} = 3.32\). ✓
(c) \(\ln 5 = 2.4 - 0.3x\), so \(x = \dfrac{2.4 - \ln 5}{0.3} = 2.64\). ✓

Analysis

[5 marks]

Some experimental values are shown.

\(x\)	1	2	3	4	5
\(y\)	2.50	7.07	12.99	20.00	27.95

It is thought that \(y = kx^n\). By considering \(\ln y\) and \(\ln x\), show that the data fit this model and find \(k\) and \(n\).

\(\ln x = 0,\ 0.693,\ 1.099,\ 1.386,\ 1.609\); \(\ln y = 0.916,\ 1.956,\ 2.564,\ 2.996,\ 3.330\). ✓
Gradients between successive points: \(\dfrac{1.040}{0.693} = 1.50\), \(\dfrac{0.608}{0.406} = 1.50\), \(\dfrac{0.432}{0.287} = 1.50\), \(\dfrac{0.334}{0.223} = 1.50\): constant, so the points lie on a straight line. ✓
So \(n = 1.5\), and the intercept (at \(\ln x = 0\)) is \(0.916 = \ln k\). ✓
\(k = e^{0.916} = 2.50\), so \(y = 2.5x^{1.5}\). ✓

Analysis

[4 marks]

The variables satisfy \(y = Ab^x\). When \(\ln y\) is plotted against \(x\), the line passes through \((1, 2.0)\) and \((3, 3.4)\).

(a) Find \(A\) and \(b\). (b) Hence find the value of \(x\) for which \(y = 100\).

(a) Gradient \(\dfrac{3.4 - 2.0}{3 - 1} = 0.7 = \ln b\), so \(b = e^{0.7} = 2.01\). Intercept \(2.0 - 0.7 = 1.3 = \ln A\), so \(A = e^{1.3} = 3.67\) (3 s.f.). ✓
(b) \(\ln 100 = 1.3 + 0.7x\), so \(x = \dfrac{\ln 100 - 1.3}{0.7} = \dfrac{4.6052 - 1.3}{0.7} = 4.72\) (3 s.f.). ✓

Analysis

[7 marks]

The number of bacteria, \(N\), in a culture \(t\) hours after it is started is modelled by \(N = Ab^t\). A graph of \(\lg N\) against \(t\) is a straight line passing through \((2, 3.10)\) and \((6, 4.30)\).

(a) Find the values of \(A\) and \(b\).

(b) Estimate the number of bacteria after 10 hours.

(a) \(\lg N = \lg A + t\lg b\). Gradient \(\dfrac{4.30 - 3.10}{6 - 2} = 0.3 = \lg b\), so \(b = 10^{0.3} = 2.00\) (3 s.f.). ✓
Intercept: \(3.10 = 0.3(2) + \lg A\), so \(\lg A = 2.5\) and \(A = 10^{2.5} = 316\) (3 s.f.). ✓
(b) \(\lg N = 2.5 + 0.3(10) = 5.5\), so \(N = 10^{5.5} \approx 316\,000\). ✓
(c) One million is \(10^6\), so \(\lg N = 6\): \(6 = 2.5 + 0.3t\). ✓
\(t = \dfrac{3.5}{0.3} = 11.7\) hours (3 s.f.). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.