Straight lines hiding in curved data
Tonight's idea
A logarithm is an exponent, and it turns multiplying into adding (where you met this). That is enough to straighten two families of curves. On this sheet \(\log\) means \(\log_{10}\).
Exponential growth, \(y = A \cdot b^x\). Take logs of both sides:
\[\log y = \log A + x \log b.\]
Call the left side \(Y\). Then \(Y = (\log b)\,x + \log A\), a straight line in \(x\). Its slope is \(\log b\) and its intercept is \(\log A\).
Power law, \(y = A x^k\). Logs again: \(\log y = \log A + k \log x\). This time plot \(\log y\) against \(\log x\). The slope of the line is \(k\), the power itself.
Log on the \(y\)-axis only (semi-log): exponentials become lines, slope \(\log b\).
Log on both axes (log-log): power laws become lines, slope \(k\).
Why bother? By eye, \(2^x\) and \(x^3\) are both curves that go up. A straight line is something you can test with a ruler, and its slope and intercept hand you the formula.
Watch one: a dish of bacteria
At \(t = 0, 2, 4, 6\) hours the counts are \(N = 50,\ 113,\ 253,\ 570\). Their logs are \(1.699,\ 2.053,\ 2.403,\ 2.756\).
Each 2-hour step adds about 0.35, so the points lie on a line and the growth is exponential. The slope is \((2.756 - 1.699) \div 6 \approx 0.176\) per hour, so \(b = 10^{0.176} \approx 1.5\). The intercept is \(1.699 = \log 50\). So \(N \approx 50 \cdot 1.5^{\,t}\): the count rises by half every hour.
Switch each axis between linear and log and watch data straighten: mathdetours.com/demos/logaxes.html.
Your turn
- For each, say which axes give a straight line (semi-log or log-log) and give its slope: \(y = 4 \cdot 3^x\), \(\;y = 5x^2\), \(\;y = 2/x\), \(\;y = 7 \cdot (0.5)^x\).
- On semi-log axes (\(\log y\) against \(x\)) a line passes through \((0, 1)\) and \((4, 3)\). Write \(y\) as a formula in \(x\).
- A pendulum of length \(L\) metres takes \(T\) seconds for one full swing. Lengths \(0.25,\ 1,\ 4\) give times \(1.00,\ 2.01,\ 4.01\) (calculated from the physics, not measured). Assume \(T = A L^k\). Find the log-log slope \(k\) from the first and last points, then find \(A\). What happens to \(T\) when the pendulum is made four times as long?
- A table gives \(x = 1, 2, 3, 4, 5\) and \(y = 2, 16, 54, 128, 250\). Without plotting, decide whether \(y\) is exponential or a power law, explain how you can tell, and find the formula.
Stretch: why exponentials always win
On semi-log axes \(y = 2^x\) is a line. What does \(y = x^5\) become, and what shape is it? Use the picture to explain why a growing exponential must eventually overtake any power. Then find the first whole number \(x > 1\) with \(2^x > x^5\).
Answers
- \(4 \cdot 3^x\): semi-log, slope \(\log 3 \approx 0.477\). \(5x^2\): log-log, slope 2. \(2/x = 2x^{-1}\): log-log, slope \(-1\). \(7 \cdot (0.5)^x\): semi-log, slope \(\log 0.5 \approx -0.301\) (a falling line: decay).
- Slope \(\tfrac{3 - 1}{4} = 0.5\), intercept 1, so \(\log y = 1 + 0.5x\) and \(y = 10 \cdot \left(10^{0.5}\right)^x \approx 10 \cdot 3.16^x\).
- \(k = \dfrac{\log 4.01 - \log 1.00}{\log 4 - \log 0.25} = \dfrac{0.603}{1.204} \approx 0.50\). At \(L = 1\), \(T = A\), so \(A \approx 2.01\) and \(T \approx 2.01\sqrt{L}\). Four times the length doubles the time, since \(4^{0.5} = 2\).
- For an exponential, equal steps in \(x\) multiply \(y\) by the same factor each time. Here the factors are \(8,\ 3.4,\ 2.4,\ 2.0\): not constant, so not exponential. For a power law, multiplying \(x\) by a fixed factor multiplies \(y\) by a fixed factor. Doubling \(x\) from 1 to 2 and from 2 to 4 both multiply \(y\) by 8, so \(2^k = 8\) and \(k = 3\). With \(y = 2\) at \(x = 1\): \(y = 2x^3\), which fits all five points.
Stretch. \(y = x^5\) becomes \(Y = 5\log x\): a log curve, which keeps rising but flattens, its slope shrinking towards zero. The line \(Y = x\log 2\) keeps the fixed slope 0.301, so it must end up above. The crossing is between 22 and 23: \(2^{22} = 4{,}194{,}304 \lt 22^5 = 5{,}153{,}632\), but \(2^{23} = 8{,}388{,}608 > 23^5 = 6{,}436{,}343\). So \(x = 23\), and \(2^x\) stays ahead from then on.