AI HL · 12.6
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Logarithmic scales on graphs
12.6.4
Screenshot and video to come
Screenshot and video to come
Screenshot and video to come
Choose a set of data, then change the axes. One choice of axes puts the points on a straight line. Predict which before you press it.
Semi-log means log y against x. Log-log means log y against log x. The logs are to base 10.
Choose the axes.
The body masses of six animals are 0.02 kg, 0.5 kg, 5 kg, 70 kg, 500 kg and 4000 kg. (a) Write the log10 of each mass. (b) Say why the logs are easier to plot than the masses.
Have a go first. Then open the working one line at a time.
From the calculator: −1.70, −0.301, 0.699, 1.85, 2.70, 3.60
Use the log key. Round to 3 significant figures.
Masses on the top row, their logs on the bottom row.
Masses run from 0.02 to 4000. Logs run from −1.70 to 3.60.
On an axis from 0 to 4000 the small masses all sit on top of 0. The logs are spread over about 5 units, and each step of 1 means a factor of 10.
The number of bacteria N (thousands) in a culture is measured every hour. (a) Plot log10 N against t and fit a straight line with the calculator. (b) Write the model in the form \(N=a\times b^{t}\).
Have a go first. Then open the working one line at a time.
log10 N: 0.556, 0.735, 0.969, 1.15, 1.37, 1.56
Keep t as it is. Only y is logged: a semi-log graph.
From the calculator: gradient \(m=0.203\), intercept \(c=0.549\), \(r=0.9995\)
r is close to 1, so the points are close to a straight line: the data is exponential.
\(b=10^{m}=10^{0.203}=1.60\), \(a=10^{c}=10^{0.549}=3.54\)
\(\log y=\log a+x\log b\): the gradient is \(\log b\) and the intercept is \(\log a\).
\(N=3.54\times1.60^{t}\)
Thousands of bacteria after t hours.
The energy use R (watts) of five animals of body mass m (kg) is recorded. (a) Plot log10 R against log10 m and fit a straight line with the calculator. (b) Write the model in the form \(R=a\,m^{k}\).
Have a go first. Then open the working one line at a time.
log10 m: −0.301, 0.301, 1.00, 1.70, 2.48 log10 R: 0.782, 1.21, 1.76, 2.26, 2.86
Both x and y are logged: a log-log graph.
From the calculator: gradient \(=0.750\), intercept \(=1.00\), \(r=0.9999\)
r is close to 1: the data follows a power law.
\(k=0.750\), \(a=10^{1.00}=10.0\)
\(\log y=\log a+k\log x\): the gradient is k and the intercept is \(\log a\).
\(R=10.0\,m^{0.750}\)
R in watts, m in kilograms.
Data are recorded for x and y. (a) Decide whether log y against x or log y against log x is a straight line. (b) Write the model, and find the exponent.
Have a go first. Then open the working one line at a time.
log10 x: 0, 0.301, 0.602, 0.903, 1.20 log10 y: 0.477, 1.08, 1.68, 2.28, 2.89
Try both graphs. Doubling x always adds 0.301 to log x.
log y against log x: \(r=1.000\). log y against x: \(r=0.9333\).
A perfect straight line has r = 1. The log-log graph is the straight one, so the relationship is a power.
Gradient \(=2.00\), intercept \(=0.477\), so \(k=2\) and \(a=10^{0.477}=3\)
Check: log 3 = 0.477.
\(y=3x^{2}\)
The graph shows \(\log_{10}y\) against x for a model \(y=a\times b^{x}\). The line passes through (0, 1.2) and (4, 2.2). (a) Find a and b. (b) Find the value of x when \(y=1000\).
Have a go first. Then open the working one line at a time.
\(m=\dfrac{2.2-1.2}{4-0}=0.25\)
The gradient of a semi-log line is \(\log b\).
\(c=1.2\)
The line meets the vertical axis at \(\log a\).
\(b=10^{0.25}=1.78\), \(a=10^{1.2}=15.8\)
The model is \(y=15.8\times1.78^{x}\).
\(\log_{10}1000=3=1.2+0.25x\), so \(x=7.20\)
Work in log form: it is a linear equation.
The graph shows \(\log_{10}y\) against \(\log_{10}x\) for a model \(y=a\,x^{k}\). The line passes through (0, 0.8) and (2, 1.8). (a) Find a and k. (b) Find y when \(x=250\).
Have a go first. Then open the working one line at a time.
\(k=\dfrac{1.8-0.8}{2-0}=0.5\)
The gradient of a log-log line is the exponent k.
\(\log a=0.8\), so \(a=10^{0.8}=6.31\)
The line meets the vertical axis where log x = 0, that is x = 1.
\(y=6.31\,x^{0.5}\)
\(y=6.31\times250^{0.5}=99.8\)
Put x = 250 into the model, not into the log graph.