Maths › Exponentials and logarithms › Log graphs and exponential models
Log graphs and exponential models
Real data rarely announces its own formula. Plot it on log axes and power laws and exponentials both confess, each becoming a straight line whose gradient and intercept hand over the constants. After that the model gets used, and eventually the model gets criticised, because every one of them breaks somewhere.
Builds on Logarithms and their laws and Straight lines.
Where it earns its keep: The time constant and exponential decay on InkPhysics.
IN THIS TOPIC
- Linearise y = axn and y = kbx with the right choice of log plot.
- Read the constants of a model off a log graph's gradient and intercept.
- Interpret, use and criticise exponential growth and decay models.
COMMON MISCONCEPTION
A straight line on a log graph means the data is linear.
Two shapes, two log plots
Take logs of a power law y = axn and last lesson's laws give log y = log a + n log x. Plot log y against log x and that is a straight line, gradient n and intercept log a. Take logs of an exponential y = kbx instead and you get log y = log k + x log b, a straight line when log y is plotted against x itself. The choice of horizontal axis is the diagnosis. Whichever plot straightens the data names the family the data belongs to.
The line lives on logged axes, so the relationship underneath is anything but linear, and straightness up there is evidence of a power law or an exponential. That is the entire point of drawing it.
WORKED EXAMPLE
Constants from a log-log line
A quantity follows y = axn, and measurements give (2, 12) and (5, 75). Find n and a.
The gradient of the log-log line is n = (log 75 − log 12)/(log 5 − log 2) = 0.7959/0.3979 = 2.
Then log a = log 12 − 2 log 2 = log 3, so a = 3, and the law is y = 3x2.
A third data point, where one is given, is there for checking, and examiners include one more often than not. Here 3 × 42 = 48 would confirm a reading at (4, 48).
GUIDED PRACTICE
Constants from a log-linear line
Plotting log10 y against x gives a straight line with intercept 0.30 and gradient 0.15. Find the model in the form y = kbx, before opening the working.
Show the working
The intercept is log k, so k = 100.30 = 2.0 to 2 significant figures.
The gradient is log b, so b = 100.15 = 1.41.
The model is y ≈ 2.0 × 1.41x, growing about 41% per unit of x. Un-logging the intercept is the step most often skipped, and log k is not yet k.
Decay, half-life and the long run
Exponential decay is the same model with a negative constant, m = Ae−kt, and it has a property no other curve has. Equal time steps multiply by equal factors, so the time taken to halve is the same wherever on the curve you start.
WORKED EXAMPLE
A decay model, read in full
A mass in grams is modelled by m = 80e−0.05t, t in days. State the initial mass, find m after 10 days, and find how long the mass takes to halve.
At t = 0 the mass is the front constant, 80 g.
m(10) = 80e−0.5 = 48.5 g.
Halving needs e−0.05t = ½, so t = ln 2/0.05 = 13.9 days, and the same 13.9 days halves it again after that.
INDEPENDENT PRACTICE
Where the model breaks
A bacteria population is modelled by P = 2000e0.1t, t in hours. Evaluate the model's prediction at t = 100, and explain why the model must fail long before then.
Show the working
P(100) = 2000e10 ≈ 4.4 × 107, twenty-two thousand times the starting population.
The model assumes the growth rate stays proportional to P forever. Food, space and waste all cap a real colony, and past some population that assumption simply stops being true.
The expected answer names the assumption, names the resource that runs out, then proposes the refinement: a model whose growth slows as P approaches a ceiling. “The number is too big” on its own is not an evaluation.
ASSESSMENT FOCUS
- Match the plot to the suspect. log y against log x for a power law, log y against x for an exponential, and say in words which you have chosen.
- The logged line's gradient and intercept are n and log a, or log b and log k. Un-log the intercept before you quote the constant.
- Half-life questions all reduce to e−kt = ½. Solve with ln and quote t = ln 2/k.
- “Initial” always means t = 0, where the exponential factor equals 1 and the front constant is your answer.
CHECK YOURSELF
The model y = 4 × 2x is to be drawn as a straight line. State what should be plotted, and give the line's gradient and intercept.
Show a hint
Take log base 10 of both sides and read the structure.
Show the answer
Taking logs gives log y = log 4 + x log 2, so plot log y against x.
The gradient is log 2 ≈ 0.301 and the intercept is log 4 ≈ 0.602.
The intercept is twice the gradient here, because 4 = 22. A small internal check the numbers happily pass.
Power laws straighten on log-log axes, exponentials on log-linear; the straightening plot is the diagnosis.
In Ae to the kt, A is the start, k the proportional rate, and ln 2 over k the halving or doubling time.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the log graphs and exponential models questions page.
CHECK YOUR PROGRESS
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- Linearise y = axn and y = kbx with the right choice of log plot.
- Read the constants of a model off a log graph's gradient and intercept.
- Interpret, use and criticise exponential growth and decay models.
Open the full revision checklist to see every objective in the course in one place.