Taking logarithms turns the curves y = axⁿ and y = kbˣ into straight lines, so the constants can be read from a graph. For y = axⁿ, plot log y against log x: the gradient is n and the intercept is log a. For y = kbˣ, plot log y against x: the gradient is log b and the intercept is log k. Raise 10 to the intercept, and for the exponential law to the gradient as well, to find the constants.
Linear forms
- y = axⁿ gives log y = n log x + log a: plot log y against log x
- Gradient n; intercept log a, so a = 10^(intercept)
- y = kbˣ gives log y = (log b)x + log k: plot log y against x
- Gradient log b, so b = 10^(gradient); intercept log k, so k = 10^(intercept)
Why taking logs gives a straight line
Both laws become linear after taking logs to base 10. For y = axⁿ, log y = log a + n log x, which has the form Y = mX + c with Y = log y and X = log x. For y = kbˣ, log y = log k + x log b, which is a straight line when log y is plotted against x itself.
The horizontal axis tells you which model you have: log x on the axis means a power law; plain x means an exponential law.
Finding a and n for a power law
On a graph of log y against log x, the gradient is n and the vertical intercept is log a. To find a, raise 10 to the power of the intercept: a = 10^(intercept). Forgetting this step, and giving the intercept itself as a, is the single most common error.
Finding k and b for an exponential law
On a graph of log y against x, both constants are logged. The gradient is log b, so b = 10^(gradient), and the intercept is log k, so k = 10^(intercept). Both values must be un-logged before you write the model.
Reading values from a graph
Read the intercept where the line meets the vertical axis. Find the gradient from two points you can read exactly, as the change in log y divided by the change in the horizontal variable. If the intercept is off the graph, use one point and the gradient to find it, as with any straight line.
Using the fitted model
Once you have y = axⁿ or y = kbˣ, use it to predict: substitute x to find y. To find x for a given y, rearrange; an exponential law needs logs again. Comment on reliability if the prediction is far outside the range of the data, which is extrapolation.
Worked examples
Example 1. Show that y = axⁿ gives a straight line when log y is plotted against log x, and state what the gradient and intercept represent.
- Take logs: log y = log(axⁿ) = log a + log xⁿ.
- Power law: log y = n log x + log a.
- This is Y = mX + c with X = log x.
Answer: Gradient n, intercept log a.
Marks: M1 for log y = log a + n log x, A1 for the gradient and intercept.
Example 2. A graph of log y against log x is a straight line with gradient 2 and intercept 0.7. Find n and a, and write the model.
- Gradient 2, so n = 2.
- Intercept 0.7 = log a, so a = 10^0.7 = 5.011…
Answer: y = 5.01x² (3 s.f.).
Marks: B1 for n = 2, M1 for a = 10^0.7, A1 for 5.01.
Example 3. A graph of log y against x passes through (0, 1) and (5, 3). Write y in the form kbˣ.
- Gradient = (3 − 1)/(5 − 0) = 0.4 = log b, so b = 10^0.4 = 2.511…
- Intercept 1 = log k, so k = 10.
Answer: y = 10 × 2.51ˣ (3 s.f.).
Marks: M1 for b = 10^0.4, B1 for k = 10, A1 for the model.
Example 4. A graph of log y against x passes through (2, 1.3) and (6, 3.3). Find the equation of the line, then b and k for y = kbˣ.
- Gradient = (3.3 − 1.3)/(6 − 2) = 0.5.
- Through (2, 1.3): 1.3 = 0.5 × 2 + c, so c = 0.3.
- log y = 0.5x + 0.3, so b = 10^0.5 ≈ 3.16 and k = 10^0.3 ≈ 2.00.
Answer: log y = 0.5x + 0.3; b ≈ 3.16, k ≈ 2.00.
Marks: M1 for the gradient, M1 for c, A1 for b and k.
Example 5. Using the model y = 5x², find y when x = 4, and x when y = 45.
- y = 5 × 4² = 80.
- 45 = 5x², so x² = 9 and x = 3, taking the positive value.
Answer: y = 80; x = 3.
Marks: A1 for each.
Common mistakes
- Giving the intercept as the constant. The intercept is log a or log k; raise 10 to it.
- Using the gradient as b for an exponential law. The gradient is log b, so b = 10^(gradient).
- Mixing up the two models. log x on the horizontal axis means y = axⁿ; plain x means y = kbˣ.
- Reading points inaccurately. Choose points where the line crosses gridlines exactly.
- Rounding before un-logging. Keep the exact gradient and intercept until you calculate the constants.
How log graphs in linear form are examined
Edexcel International A-Level tests this on the Pure 3 paper (WMA13); in Edexcel A-Level (9MA0) it is taught in Year 1. Nearly every question gives a graph:
- explain why plotting logs gives a straight line
- find the equation of the line from marked points
- find the constants a and n, or k and b, to 3 significant figures
- use the model to estimate a value, and comment on reliability
For graph-reading practice with full marking, use the Pure 3 logarithms exam questions. The Pure 3 exponentials and logarithms worksheet and the logarithms revision engine give extra practice.
You need the laws of logarithms and straight line equations. This sits alongside exponential models.
Frequently asked questions
How do you know whether to plot log y against log x or against x?
For a power law y = axⁿ, plot log y against log x. For an exponential law y = kbˣ, plot log y against x.
How do you find a from the intercept?
The intercept is log a, so a = 10 raised to the power of the intercept.
Why use logs to turn a curve into a straight line?
A straight line is easy to fit to data, and its gradient and intercept give the constants of the original model.