Papers 2 and 3. Taking logarithms turns \(y = kx^n\) and \(y = ka^x\) into straight-line graphs. Read the gradient and intercept of the line, then undo the logs to find the unknown constants.
\(y = kx^n \Rightarrow \ln y = \ln k + n\ln x\): plot \(\ln y\) against \(\ln x\); gradient \(n\), intercept \(\ln k\). \(y = ka^x \Rightarrow \ln y = \ln k + x\ln a\): plot \(\ln y\) against \(x\); gradient \(\ln a\), intercept \(\ln k\). Compare with \(Y = mX + c\).
Fig. 1: for y = A b^x, the graph of ln y against x is a straight line. Here it passes through (1, 1.6) and (4, 2.8), so the gradient ln b = 0.4 and the intercept ln A = 1.2
Section 01
Which graph is a straight line?
Take ln of both sides of the model and expand with the laws of logarithms. Then match the result, term by term, with \(Y = mX + c\): whatever multiplies the constant gradient is your \(X\), and the log of the outcome is your \(Y\).
Model
Take ln
Plot \(Y\) against \(X\)
Gradient
Intercept
\(y = kx^n\)
\(\ln y = n\ln x + \ln k\)
\(\ln y\) against \(\ln x\)
\(n\)
\(\ln k\)
\(y = ka^x\)
\(\ln y = (\ln a)x + \ln k\)
\(\ln y\) against \(x\)
\(\ln a\)
\(\ln k\)
\(y = Ae^{bx}\)
\(\ln y = bx + \ln A\)
\(\ln y\) against \(x\)
\(b\)
\(\ln A\)
Worked example Choosing the axes
The variables satisfy \(y = 5x^{3}\). Show that a suitable graph is a straight line, and state its gradient and intercept.
Take ln: \(\ln y = \ln(5x^3) = \ln 5 + \ln x^3\).
Power law: \(\ln y = 3\ln x + \ln 5\).
This is \(Y = mX + c\) with \(Y = \ln y\), \(X = \ln x\), \(m = 3\), \(c = \ln 5\).
Answer: \(\ln y\) against \(\ln x\): a straight line with gradient \(3\) and intercept \(\ln 5 \approx 1.61\)
Worked example Undoing a line equation
The graph of \(\ln y\) against \(\ln x\) is the straight line \(\ln y = 0.8\ln x + 1.5\). Express \(y\) in terms of \(x\).
Write \(1.5 = \ln e^{1.5}\) and \(0.8\ln x = \ln x^{0.8}\).
\(\ln y = \ln\left(e^{1.5}x^{0.8}\right)\), so \(y = e^{1.5}x^{0.8}\).
For \(y = ka^x\) the horizontal axis is plain \(x\), not \(\ln x\): the \(x\) is already in the index, and taking logs brings it down as it is. Plot \(\ln x\) only when \(x\) is raised to a constant power.
Section 02
Constants from two points.
Exam questions give two points on the straight line (or one point and the gradient). Find the gradient, then the intercept, then exponentiate: the gradient and intercept are logarithms of the constants, not the constants themselves.
Worked example Exponential model, ln y against x
The variables satisfy \(y = Ab^x\). The graph of \(\ln y\) against \(x\) is a straight line through \((1, 1.6)\) and \((4, 2.8)\). Find \(A\) and \(b\).
Gradient: \(\dfrac{2.8 - 1.6}{4 - 1} = \dfrac{1.2}{3} = 0.4\), so \(\ln b = 0.4\).
The variables satisfy \(y = kx^n\). The graph of \(\ln y\) against \(\ln x\) is a straight line through \((0.5, 2.3)\) and \((2.0, 1.4)\). Find \(k\) and \(n\).
The coordinates in these questions are values of \(\ln y\) (and perhaps \(\ln x\)), not of \(y\). The intercept of the line is \(\ln k\); answering \(k = 2.6\) instead of \(k = e^{2.6}\) loses the final marks.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
For the model \(y = kx^n\), which graph is a straight line?
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For \(y = ka^x\), the gradient of the graph of \(\ln y\) against \(x\) is
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For \(y = Ae^{bx}\), the graph of \(\ln y\) against \(x\) has intercept 2. What is \(A\)?
Quick check+10 XP
If \(\ln y = 3 + 2\ln x\), then \(y\) equals
Section 03
From a table of data.
Experimental data rarely come labelled with their model. Work out the logarithms, check that they lie on (or close to) a straight line, and use the line to find the constants. Base 10 (lg) works in exactly the same way; just undo with \(10^{(\dots)}\) instead of \(e^{(\dots)}\).
Worked example Confirming an exponential model
Measurements give \(x = 1, 2, 3, 4\) and \(y = 6.0, 10.8, 19.4, 35.0\). Show that \(y \approx Ab^x\) and estimate \(A\) and \(b\).
\(\ln y = 1.792,\ 2.380,\ 2.965,\ 3.555\) (3 d.p.).
Successive differences are \(0.588, 0.585, 0.590\): almost constant, so \(\ln y\) against \(x\) is (very nearly) a straight line, and the model fits.
Gradient (first and last points): \(\dfrac{3.555 - 1.792}{3} = 0.588 = \ln b\), so \(b = e^{0.588} = 1.80\).
Undo with the SAME base you took logs in. If the axis is \(\lg y\), use \(10^{1.2}\); using \(e^{1.2}\) gives \(3.32\), a completely different (wrong) constant.
Stage 2 · Exam
Exam-style questions
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\(y = Ab^x\). The graph of \(\ln y\) against \(x\) is a straight line through \((2, 3.1)\) and \((5, 4.6)\). Find \(A\) and \(b\), to 3 significant figures.
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\(y = kx^n\). The graph of \(\ln y\) against \(\ln x\) is a straight line through \((1, 4)\) and \((3, 1)\). Find \(k\) and \(n\).
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\(y = Ab^x\). The graph of \(\lg y\) against \(x\) has gradient \(0.3\) and intercept \(1.2\). Find \(A\) and \(b\).
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The graph of \(\ln y\) against \(\ln x\) is the line \(\ln y = 0.8\ln x + 1.5\). Find \(y\) when \(x = 10\).
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Straight lines, from curves.
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