EL4, EL5 - Straight-line graphs for power laws

Distinguish a power law from an exponential model, then use logarithms on both variables to make a straight line. EL4 gives the linear form of a power law; EL5 uses the fitted graph to estimate its constant and power. You will interpret those estimates and test the limits of predictions.

Taking logarithms of both variables

If doubling the edge of a cube multiplies its volume by $8$ , how could a straight-line graph reveal the power $3$ ? A plot of volume against edge length is curved, but a plot of their logarithms makes the power a gradient.

You already know

For $y=ka^x$ , plotting $\log_{10}y$ against $x$ gives a straight line. The product law splits a logarithm of a product; the power law brings an exponent outside as a multiplier. A fitted line $Y=mX+c$ has gradient $m$ and intercept $c$ .

A power law has the form $y=kx^n$ , where $k$ and $n$ are constants: the variable is the base, not the exponent. Examples include circle area proportional to radius squared and cube volume proportional to edge length cubed. Doubling a cube's edge gives $(2x)^3=8x^3$ , regardless of the starting size. For logarithmic graphs we use $x>0$ and $k>0$ , so both plotted logarithms exist; $n$ may be positive, negative or zero.

Linear form of a power law

\begin{aligned} y&=kx^n\\ \log_{10}y&=\log_{10}k+n\log_{10}x \end{aligned}

Plot $Y=\log_{10}y$ against $X=\log_{10}x$ . Then the gradient is $n$ , the intercept is $\log_{10}k$ , and $k=10^c$ . Here $x,k>0$ , and $n$ is real.
Not on the formulae sheet: learn it.

The critical difference from the previous lesson is the horizontal coordinate: now take its logarithm too.

Worked example 1: a positive power

Question. Show that $y=5x^3$ , for $x>0$ , gives a straight line on a graph of $\log_{10}y$ against $\log_{10}x$ . State its gradient and intercept.

Plan. Take logarithms, apply both laws, then match the transformed variables to a straight line.

\begin{aligned} \log_{10}y&=\log_{10}(5x^3) && \text{take logs}\\ &=\log_{10}5+\log_{10}(x^3) && \text{product law}\\ &=\log_{10}5+3\log_{10}x && \text{power law} \end{aligned}

Why is the gradient $3$ ? It multiplies the horizontal coordinate $\log_{10}x$ , exactly as $m$ multiplies $X$ .

Answer. Gradient $3$ , intercept $\log_{10}5\approx0.699$ .

Check. At $x=1$ , the horizontal coordinate is $0$ and $y=5$ , so the vertical coordinate is $\log_{10}5$ .

Change only the sign of the power. The line now slopes down, while its intercept stays the same.

Worked example 2: a negative power

Question. Show that $y=5x^{-3}$ , for $x>0$ , gives a straight line on the same transformed axes. State its gradient and intercept.

Plan. Apply the same laws, keeping the negative exponent with its coefficient.

\begin{aligned} \log_{10}y&=\log_{10}(5x^{-3}) && \text{take logs}\\ &=\log_{10}5+\log_{10}(x^{-3}) && \text{product law}\\ &=\log_{10}5-3\log_{10}x && \text{power law} \end{aligned}

Why is the logarithm allowed? The exponent is negative, but $5x^{-3}=5/x^3$ remains positive for positive $x$ .

Answer. Gradient $-3$ , intercept $\log_{10}5$ .

Check. Doubling $x$ divides $y$ by $8$ , so the log-height changes by $-3\log_{10}2$ , matching the gradient.

A zero power gives a horizontal transformed line: $y=kx^0=k$ , so $\log_{10}y=\log_{10}k$ . In all three cases, transformed horizontal coordinate zero means $x=1$ , not $x=0$ .

The logarithmic plot always excludes $x=0$ . A negative-power model itself also excludes zero: for example, $5x^{-3}=5/x^3$ would divide by zero there, even before taking logarithms.

Watch out

Wrong: plot $\log_{10}y$ against $x$ for $y=kx^n$ . The linear expression contains $\log_{10}x$ . Right: label both axes with their logarithms; the gradient is $n$ , not $\log_{10}n$ .

Keep the negative exponent, but change its magnitude to a fraction.

Estimating the power and constant from a graph

Transform both rows of a data table before plotting, then estimate a line through the middle of the scatter.

For example, inputs $1, 2, 4, 8, 16$ have evenly spaced logarithms $0, 0.301, 0.602, 0.903, 1.204$ . The original inputs are not equally spaced, but their logarithms are. Your axis scale must describe the quantities you actually plot.

Worked example 3: a rising power-law plot

Question. Estimate $k$ and $n$ in $y=kx^n$ from this dataset and its transformed plot.

$x$	$1$	$2$	$4$	$8$	$16$
$y$	$3.2$	$8.8$	$25.6$	$70.8$	$204$
$\log_{10}x$	$0$	$0.301$	$0.602$	$0.903$	$1.204$
$\log_{10}y$	$0.505$	$0.944$	$1.408$	$1.850$	$2.310$

Plan. Plot the paired logarithms, read a large gradient triangle from the fitted line and undo the intercept's logarithm.

The left panel below shows the points and a suitable fitted line. Two convenient points on the line are approximately $(0, 0.50)$ and $(1.20, 2.30)$ .

\begin{aligned} n=m&\approx\frac{2.30-0.50}{1.20-0}=1.50 && \text{gradient of the line}\\ c&\approx0.50 && \text{vertical intercept}\\ k&\approx10^{0.50}=3.1623\ldots && \text{undo logarithm} \end{aligned}

Why does $n$ need no power of $10$ ? It is already the line's gradient; only $k$ is hidden inside a logarithm.

Answer. $n\approx1.50$ , $k\approx3.16$ , so $y\approx3.16x^{1.50}$ . The near-straight scatter supports this model over the observed range.

Check. At $x=4$ , the fitted equation gives approximately $3.1623\times8=25.3$ , close to $25.6$ .

Now change the trend from increasing to decreasing, keeping the same horizontal inputs and plotting method. A negative power is read directly from the falling line.

Worked example 4: a falling power-law plot

Question. Estimate $k$ and $n$ in $y=kx^n$ from the second dataset and the right-hand transformed graph.

$x$	$1$	$2$	$4$	$8$	$16$
$y$	$31.8$	$11.0$	$4.0$	$1.38$	$0.50$
$\log_{10}x$	$0$	$0.301$	$0.602$	$0.903$	$1.204$
$\log_{10}y$	$1.502$	$1.041$	$0.602$	$0.140$	$-0.301$

Plan. Use two separated points on the fitted line, then recover the positive coefficient from its intercept.

Read approximately $(0, 1.50)$ and $(1.20, -0.30)$ .

\begin{aligned} n&\approx\frac{-0.30-1.50}{1.20}=-1.50 && \text{fall divided by run}\\ \log_{10}k&\approx1.50 && \text{read intercept}\\ k&\approx10^{1.50}=31.6228\ldots && \text{undo logarithm} \end{aligned}

Why may a plotted vertical coordinate be negative? It represents a positive value below $1$ , not a negative original measurement.

Answer. $n\approx-1.50$ , $k\approx31.6$ , so $y\approx31.6x^{-1.50}$ .

Check. At $x=16$ , $31.6228/16^{1.5}\approx0.494$ , close to $0.50$ .

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Algebra

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Calculus

Exponentials and logarithms

Numerical methods

EL4, EL5 - Straight-line graphs for power laws

Taking logarithms of both variables

Estimating the power and constant from a graph

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