Maths › Exponentials and logarithms › Log graphs and exponential models
Log graphs and exponential models
Real data rarely announces its formula. Plot it on log axes, though, and power laws and exponentials both confess, each becoming a straight line whose gradient and intercept hand over the constants. Then the model does what models do: predicts, decays, compounds, and eventually breaks.
Builds on Logarithms and their laws and Straight lines.
IN THIS TOPIC
- Linearise y = axn and y = kbx with the right choice of log plot.
- Read the constants of a model from a log graph's gradient and intercept.
- Interpret, use and criticise exponential growth and decay models.
WHAT YOU PROBABLY THINK
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 the laws from last lesson give log y = log a + n log x, which is a straight line when log y is plotted against log x, gradient n, intercept log a. Take logs of an exponential y = kbx and instead 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 it belongs to.
That settles the opening lie. The line lives on logged axes, so the relationship underneath is anything but linear; straightness there is evidence of a power law or an exponential, which is precisely why the plot is useful.
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, if given, is for checking, and examiners include one more often than not. 3 × 42 = 48 would confirm a point at (4, 48).
YOUR TURN
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; log k is not yet k.
Decay, half-life and the long run
Exponential decay is the growth machinery with a negative constant, m = Ae−kt, and it has a signature no other curve shares. Equal time steps multiply by equal factors, so the time to halve is the same from anywhere on the curve.
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 means e−0.05t = ½, so t = ln 2/0.05 = 13.9 days, and the same 13.9 days halves it again thereafter.
The half-life computation is the logs lesson working inside the models lesson, which is how these chapters were always going to meet.
TRY IT UNSEEN
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, but food, space and waste all cap real colonies. Beyond some population the assumption fails and growth levels off.
The expected answer names the assumption and the resource limit, then suggests the refinement: a model whose growth slows as P approaches a ceiling.
THE EXAM BIT
- Choose the plot to match the suspect: log y against log x for power laws, log y against x for exponentials, and say which you are using.
- Gradient and intercept of the logged line are n and log a (or log b and log k); un-log the intercept before quoting the constant.
- Half-life questions reduce to e−kt = ½; solve with ln and quote t = ln 2/k.
- “Initial” always means t = 0, where the exponential factor is 1 and the front constant is the answer.
- Model-criticism marks want the failing assumption named and a bounded refinement suggested, never only “the number is too big”.
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: 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.
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 from a log graph's gradient and intercept.
- Interpret, use and criticise exponential growth and decay models.
No animated video for this topic yet; these notes stand alone.