Maths › Exponentials and logarithms › Log graphs and exponential models
Log graphs and exponential models
Plotting log y against log x gives a straight line for a power law, and plotting log y against x gives a straight line for an exponential, with the gradient and intercept determining the constants. Fit such a model to data, use it, and state the conditions under which it stops being reasonable.
Builds on Logarithms and their laws and Straight lines.
Used in InkPhysics: The time constant and exponential decay.
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.
The straight line is on logarithmic axes, so the relationship between x and y is not linear. A line on log y against log x indicates a power law y = axn, with gradient n and intercept log a; a line on log y against x indicates an exponential y = kbx, with gradient log b and intercept log k. The plot that straightens the data identifies the family.
Two shapes, two log plots
Taking logarithms of a power law y = axn, for x > 0 and a > 0, gives 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. Whichever plot gives a straight line identifies the model that fits the data.
The straight line is on logarithmic axes, so the relationship between x and y is not linear; a straight line on those axes is evidence of a power law or an exponential relationship.
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.
Use any third data point to check the fitted model: 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, an increase of about 41% per unit of x. The intercept is log k rather than k, so it must be un-logged.
Decay, half-life and the long run
Exponential decay is the same model with a negative constant, m = Ae−kt with k > 0. Equal time intervals multiply the quantity by equal factors, so the time taken to halve is the same from any starting point 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 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 remains proportional to P for all t. In a real colony, food, space and the accumulation of waste limit the population, so the assumption fails above some size.
A complete answer names the assumption, names the limiting resource, and proposes a specific refinement, such as a model whose growth rate decreases as P approaches a limiting value.
ASSESSMENT FOCUS
- Choose the plot to match the proposed model: log y against log x for a power law, log y against x for an exponential, and state which has been chosen.
- The gradient and intercept of the logarithmic plot are n and log a for a power law, or log b and log k for an exponential. Un-log the intercept before quoting the constant.
- A half-life question reduces to e−kt = ½, so t = ln 2/k.
- “Initial” means the value at t = 0, where the exponential factor is 1 and the value is the constant A.
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 because 4 = 22, which checks the two values.
A power law gives a straight line on log-log axes and an exponential gives one on log-linear axes; the plot that straightens the data identifies the model.
In Aekt, A is the value at t = 0, k is the proportional rate of change, and ln 2 divided by |k| is the doubling or halving 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.
USED LATER IN
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.
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