Transform power and exponential relationships to straight-line form
Take logarithms to turn power and exponential models into a straight line whose gradient and intercept reveal their constants.
Cambridge IGCSE Additional Mathematics 0606, 2028–2030 syllabus, Version 1. Paper 1 is non-calculator; Paper 2 requires a scientific calculator. These original lessons provide self-checked topic practice, not a Cambridge grade or a timed paper. Full-paper assessment and human specialist review are still outstanding.
Before you begin
- Cambridge assumes knowledge of IGCSE Mathematics or an equivalent course. Check the domain before using an inverse or logarithm, and retain exact values where the question asks for them.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which transformed axes make y = Axⁿ linear?
What you will learn
- Transform y = Axⁿ into a linear relationship using logarithms
- Transform y = Abˣ into a linear relationship using logarithms
- Read the model constants from a plotted gradient and intercept
A power law becomes linear on logarithmic axes
For y = Axⁿ with A > 0, x > 0 and y > 0, take logarithms: log y = log A + n log x. Plot log y vertically against log x horizontally.
The graph is a straight line with gradient n and vertical intercept log A. Thus n is read directly and A is the antilogarithm of the intercept in the chosen base.
An exponential law becomes linear on a semi-log plot
For y = Abˣ with A > 0, b > 0 and y > 0, take natural logarithms: ln y = ln A + x ln b.
Plot ln y vertically against x horizontally. The gradient is ln b and the intercept is ln A, so A = e^(intercept) and b = e^(gradient).
Use the same logarithm base throughout. Logs require positive x and y for the power-law transformation and positive y for the exponential transformation.
Pause and explain
For y = Abˣ, what is the gradient when plotting ln y against x?
Worked example
A graph of log₁₀ y against log₁₀ x has gradient 3 and intercept log₁₀ 5. State the relationship between x and y.
Show the worked solution
- For y = Axⁿ, the transformed gradient is n, so n = 3.
- The intercept is log₁₀ A = log₁₀ 5, so A = 5.
- Substitute the constants into the original power form.
Answer y = 5x³.
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 log₁₀ y against log₁₀ x graph has gradient 2 and intercept log₁₀ 6. Find the power model.
Give me a hint
Compare with log y = log A + n log x.
Compare my reasoning
- The gradient gives n = 2.
- The intercept gives log₁₀ A = log₁₀ 6, so A = 6.
- Write the original relationship y = Axⁿ.
y = 6x².
Look for these in your work
- Power read from gradient
- Coefficient recovered from intercept
- Original form restored
A plot of ln y against x has gradient 0.4 and intercept ln 25. Write y in the form Abˣ.
Give me a hint
The intercept gives A; exponentiate the gradient to recover b.
Compare my reasoning
- A = e^(ln 25) = 25.
- ln b = 0.4, so b = e^0.4.
- Substitute into y = Abˣ.
y = 25(e^0.4)ˣ, equivalently y = 25e^(0.4x).
Look for these in your work
- Initial factor recovered
- Base recovered from gradient
- Exponential model stated
Experimental data produce a straight line when ln y is plotted against ln x, with gradient −1 and intercept ln 12. State the power relationship and its required domain.
Give me a hint
Use the intercept for A and the gradient for n.
Compare my reasoning
- The gradient gives n = −1.
- The intercept gives A = 12.
- The log transformation requires x > 0 and y > 0.
y = 12x⁻¹ = 12/x, for x > 0.
Look for these in your work
- Power sign retained
- Coefficient recovered
- Transformation domain stated
Common mistakes
- Plotting log x against log y when the question specifies the opposite axes
- Reading the intercept itself as A instead of taking its antilogarithm
- Using different log bases in one transformed equation
- Taking logs when a model value is not positive
What can you explain now?
In a plot of log y against log x for y = Axⁿ, what does the gradient represent?
Compare with the explanation
The power n
Taking logs gives log y = log A + n log x.
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Make it stick
For each of y = Axⁿ and y = Abˣ, name the axes to plot and explain what the gradient and intercept mean.
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.