Linearise a power model and recover its constants
Plot log y against log x, read the exponent from the gradient and recover the multiplier by undoing the intercept's logarithm.
Exam scope and assessment notes
Pure Mathematics 2 (Paper 2, AS only) and Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027. Paper 1 knowledge is assumed. The AS route using Papers 1 + 2 cannot be carried forward to A Level. Shared sections 2.2 and 3.2 include log laws excluding change of base, natural exponential and logarithmic graphs, equations and inequalities, and straight-line transformations. Written practice is self-checked.
Before you begin
- Use log product and power laws.
- Find a straight-line gradient and intercept from coordinates.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
On a plot of ln y against ln x for y = kxⁿ, what is the vertical intercept?
What you will learn
- Transform y = kxⁿ for positive x, y and k into a straight-line relation.
- Recover n and k from a gradient, intercept or supplied observations, while distinguishing fitting from model validation.
Choose the transformed horizontal and vertical variables
For y = kxⁿ with x > 0 and k > 0, taking ln gives ln y = ln k + n ln x. Define horizontal X = ln x and vertical Y = ln y; then Y = nX + ln k. The gradient is n and the vertical intercept is ln k, so k = e^(intercept). The intercept is not k itself. If using base-10 logs consistently, the gradient is still n but the intercept is log₁₀ k, which is undone by a power of 10.
Suppose the transformed plot passes through (X, Y) = (0, ln 3) and (ln 2, ln 12). Its gradient is [ln 12 − ln 3]/ln 2 = ln 4/ln 2 = 2. The intercept is ln 3, giving k = 3. Therefore the model is y = 3x². The raw observations are (x, y) = (1, 3) and (2, 12); a zero horizontal log coordinate corresponds to x = 1, not x = 0.
A fitted straight line supplies constants under an assumed model
For a transformed line ln y = −(1/2)ln x + ln 4, the constants are n = −1/2 and k = 4, giving y = 4/√x for positive x. A negative slope is permitted and describes a decreasing power relation. The graph orientation matters: plotting ln x against ln y generally changes the gradient; always label which quantity is vertical before reading a coefficient.
Two observations can determine k and n if a power model is assumed. They do not independently establish that the form fits unobserved inputs; further observations are needed to assess that. Use positive dimensionless ratios or consistently defined numerical measurements before taking logarithms, and retain the units used to define a numerical model. Zero or negative data cannot be placed on these real-log axes, and changing measurement units can change the numerical multiplier k.
Pause and explain
For ln y = 3 ln x + ln 2, what are k and n?
Worked example
A plot of ln y against ln x passes through (0, ln 3) and (ln 2, ln 12). Assuming y = kxⁿ with positive inputs, find k, n and the model.
Show the worked solution
- The gradient is n = (ln 12 − ln 3)/ln 2 = ln 4/ln 2 = 2.
- The vertical intercept is ln k = ln 3, so k = 3.
- The recovered model is y = 3x²; it reproduces raw values 3 at x = 1 and 12 at x = 2.
Answer n = 2, k = 3; model y = 3x² for x > 0.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A plot with X = ln x and Y = ln y has equation Y = 3X + ln 2. Find the power model and its value at x = 2.
Give me a hint
The slope supplies the exponent; exponentiate the intercept for the multiplier.
Compare my reasoning
- The exponent is n = 3 and ln k = ln 2, so k = 2.
- The model is y = 2x³ on positive x.
- At x = 2 it predicts y = 2(8) = 16.
y = 2x³; predicted y(2) = 16.
Look for these in your work
- Read the gradient and intercept from the correct transformed axes.
- Undo the intercept's logarithm before calculating a raw prediction.
A plot of log₁₀ y against log₁₀ x has gradient −2 and intercept log₁₀ 5. Find the model and y at x = 2.
Give me a hint
Keep the same base for both axes and undo its intercept using a power of 10.
Compare my reasoning
- The exponent is n = −2 and log₁₀ k = log₁₀ 5, so k = 5.
- The model is y = 5x^(−2) = 5/x² for x > 0.
- At x = 2 it gives y = 5/4, a positive value despite the negative exponent.
y = 5/x²; y(2) = 5/4, for x > 0.
Look for these in your work
- Use one common log base and recover a positive k.
- Interpret a negative gradient as a negative exponent.
Positive numerical observations (x, y) = (2, 12) and (6, 108) are assumed to follow y = kxⁿ. Find k and n, predict y at x = 4 and state what the two points alone cannot prove.
Give me a hint
Divide the two model equations to eliminate k, then use one observation to recover it.
Compare my reasoning
- The ratio gives 108/12 = (6/2)ⁿ, so 9 = 3ⁿ and n = 2.
- Use 12 = k(2²) to get k = 3; the model is y = 3x² and predicts y(4) = 48.
- The two points determine constants under the assumed power form; they do not independently validate that form for other observations.
n = 2, k = 3; predicted y(4) = 48; further data are needed to validate the assumed model.
Look for these in your work
- Recover both constants from the supplied observations.
- Distinguish a prediction under an assumed form from independent model validation.
Common mistakes
- Recover k by undoing the log intercept; do not use the intercept itself as k.
- For a power model, both axes are transformed with a common log base.
- Two fitted points do not independently validate the model outside those observations.
What can you explain now?
Recover the power model from ln y = −(1/2)ln x + ln 4.
Compare with the explanation
y = 4x^(−1/2) = 4/√x, for x > 0.
The gradient is −1/2 and the intercept ln 4 gives multiplier 4.
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Make it stick
Label both transformed axes for a power model, recover its gradient and multiplier, and explain why fitting two observations is different from validating the model.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.