Linearise power and exponential models
Choose transformed axes, recover power or exponential constants and test the model against original data and domain limits.
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Change the variables before reading the gradient
A curved relationship can become a straight line when we plot different variables. The gradient and intercept belong to the labelled axes, not automatically to x and y. Rewrite the proposed relationship as Y = mX + c, name X and Y, then match coefficients.
Recall that log ₁₀y = q means y = 10^q, while ln y = q means y = e^q. Taking a logarithm and undoing it are different operations. All x and y values in the examples below are dimensionless; for measured quantities, first fix the units used for their numerical values.
A power model becomes linear in two logarithms
For y = axⁿ with a > 0 and x > 0, y is positive. Taking base-ten logarithms gives
With X = log ₁₀x and Y = log ₁₀y, this is a straight line of gradient n and vertical intercept log ₁₀a. The intercept is not a itself.
Suppose the transformed graph passes through (X,Y) = (0,1), (1,3) and (2,5). Its gradient is (5 − 1)/(2 − 0) = 2 and its intercept is 1. Therefore n = 2 and a = 10¹ = 10, giving y = 10x².
| Original x | Original y | X = log₁₀x | Y = log₁₀y |
|---|---|---|---|
| 1 | 10 | 0 | 1 |
| 10 | 1000 | 1 | 3 |
| 100 | 100000 | 2 | 5 |
The transformed intercept at X = 0 corresponds to x = 1, because log₁₀1 = 0. It is not the original y-intercept at x = 0. Check the recovered model on the original scale: x = 10 gives y = 10 × 10² = 1000.
A power model is straight on two logarithmic variables
An increasing straight line on transformed coordinates, with marked points (0,1), (1,3) and (2,5). The horizontal axis is log₁₀x, rather than x, and the vertical axis is log₁₀y. Its vertical intercept is one and gradient two.
Scroll across the figure to read all labels.
View figure data
| Series | X = log₁₀x (dimensionless) | Y = log₁₀y (dimensionless) |
|---|---|---|
| Y = 1 + 2X | 0 | 1 |
| Y = 1 + 2X | 2 | 5 |
| Transformed model points | 0 | 1 |
| Transformed model points | 1 | 3 |
| Transformed model points | 2 | 5 |
Using natural logarithms instead gives ln y = ln a + n ln x. The slope is still n when both axes use the same logarithm base, but the intercept now needs a = eᶜ. For a line with gradient 1.5 and intercept 0.7 on these natural-log axes, y = e^(0.7)x^(1.5). The coefficient is e^0.7, rather than 0.7.
Check your understanding 1 Read the transformed origin correctly
A graph of ln y against ln x has gradient −1/2 and vertical intercept ln 6. Recover the power model, find y at x = 4 and explain which original x-value corresponds to that transformed intercept.
Show answer
The line is ln y = ln 6-1/2 ln x, so y = 6x^(-1/2) = 6/square root of x for x > 0. At x = 4, y = 3. The transformed intercept has ln x = 0, hence x = 1 and y = 6. It does not supply a value at x = 0, where this model is undefined. A negative slope here means a negative power, not negative y-values.
An exponential model leaves the input unlogged
For y = kb^x with k > 0 and b > 0, taking base-ten logarithms gives
Choose X = x, rather than log₁₀x, and Y = log₁₀y. The line has gradient log₁₀b and intercept log₁₀k. Both constants need an inverse logarithm: k = 10ᶜ and b = 10ᵐ. A negative gradient gives 0 < b < 1, rather than a negative base.
Suppose this graph passes through (0,2), (2,1) and (4,0). Its gradient is −1/2 and intercept 2. Thus k = 100 and b = 10^(-1/2), giving y = 100(10^(-1/2))^x.
| x, unchanged | y | Y = log₁₀y |
|---|---|---|
| 0 | 100 | 2 |
| 2 | 10 | 1 |
| 4 | 1 | 0 |
At x = 2, the model gives 100 × 10⁻¹ = 10, agreeing with the original value. Here the transformed intercept really does occur at x = 0, since the horizontal axis was not logged.
An exponential model leaves x unchanged
A decreasing straight line through marked transformed points (0,2), (2,1) and (4,0). The horizontal axis is the original input x; only the vertical variable is log₁₀y. Its gradient is minus one half and vertical intercept two.
Scroll across the figure to read all labels.
View figure data
| Series | X = x (dimensionless) | Y = log₁₀y (dimensionless) |
|---|---|---|
| Y = 2 − X/2 | 0 | 2 |
| Y = 2 − X/2 | 4 | 0 |
| Transformed model points | 0 | 2 |
| Transformed model points | 2 | 1 |
| Transformed model points | 4 | 0 |
Natural logarithms work too: ln y = ln k + x ln b. A line of gradient −0.4 and intercept 2 gives y = e²(e^(-0.4))^x, equivalently y = e^(2-0.4x). At x = 5, ln y = 0, so y = 1; substituting in the original exponential confirms it.
Check your understanding 2 Recover the base, not its logarithm
A graph of log₁₀y against x is a line with gradient 0.3 and intercept 1.2. Express y in the form kbˣ and find its value at x = 1, retaining exact powers of ten. Explain why y = 1.2(0.3)ˣ is not the recovered model.
Show answer
log ₁₀y = 1.2 + 0.3x, so k = 10^1.2 and b = 10^0.3. Thus y = 10^(1.2)(10^(0.3))^x, and at x = 1, y = 10^1.5. The proposed model incorrectly uses the logarithms as the constants. It would also decay because 0 < 0.3 < 1, whereas the given positive gradient means log₁₀b > 0 and therefore b > 1.
Check exactly which quantities need logarithms
For a power log–log plot, both x and y must be positive. For an exponential plot of ln y against x, only y is logged: x = 0 or a negative x is allowed. For instance, y = 4(1/2)ˣ gives y = 4 at x = 0 and y = 8 at x = −1; both points are valid on its ln y against x graph. Rejecting them merely because ln x does not exist would apply a transformation that this model never requested.
Neither transformation allows a logarithm of y = 0 or y < 0. Do not silently omit such observations or add an arbitrary constant to make them positive; first reconsider the proposed model or whether a different given relationship is intended. The constant cases also matter: n = 0 gives y = a, while b = 1 gives y = k. The same positive constant data can fit both forms, so straightness alone need not distinguish them.
Other given relationships may need a different substitution. For y = a/x + b with x ≠ 0, set X = 1/x and Y = y. Then Y = aX + b: the gradient is a and intercept b, without any logarithm. For example, y = 3/x + 2 gives points (X,Y) = (1,5), (1/2,7/2), (1/4,11/4), all on Y = 3X + 2. Read and rearrange the relationship rather than selecting a log plot by habit.
Test the recovered model on the original quantities
Two transformed points determine a line, but further observations test whether that line describes the data. Unequal original x-steps are particularly easy to misread: a power model has a fixed y-factor for a fixed multiplication of positive x, while an exponential model has a fixed y-factor for a fixed addition to x.
Real observations may only lie approximately on a transformed line. Use well-separated points on the chosen straight line to estimate its gradient, retain sufficient precision when undoing logarithms, and compare predictions with the original data. This establishes a useful fit, not an exact law or a guarantee outside the observed range.
Try it yourself 1
Choose a model from the data and challenge it
Problem
An ideal data set is proposed to follow either y = axⁿ or y = kbˣ, with positive constants a, k and b:
| x | 1 | 2 | 4 | 8 |
|---|---|---|---|---|
| y | 10 | 5 | 2.5 | 1.25 |
Choose a suitable straight-line plot, state its equation and recover the original model. A proposal says that halving y proves an exponential model and fits y = 20(1/2)ˣ using the first two observations. Test this claim against an unused point. Predict y at x = 5 and explain how the status of a prediction at x = 16 differs.
Write your answer, then compare it with the solution
Hints
Hint 1
Compare the input steps before interpreting the output factor
The y-values halve each time, but inspect whether the x-values increase by equal differences or equal factors. Choose variables that turn the supported relationship into Y = mX + c, then test a point that was not used to recover the constants.
Show full solution
Identify the relationship and its axes
Method
Compare the factors in x and y across the whole table.Reason
Halving y for doubled x describes a power relationship; it does not establish a constant multiplier for equal additions to x.Working
Answer
Every product xy is 10, so y = 10/x. With X = log₁₀x and Y = log₁₀y, the straight line is Y = 1 − X. For example, x = 1 gives (X,Y) = (0,1), while x = 2 gives (log₁₀2,1 − log₁₀2). The gradient is −1 and intercept 1.Recover and verify the original model
Method
Match the power-model coefficients and return to the original variables.Reason
The slope is n and intercept is log₁₀a on these axes.Working
Answer
n = −1 and a = 10¹ = 10, so y = 10x⁻¹. At x = 4 this gives 2.5, and at x = 8 it gives 1.25, matching the unused observations. A valid alternative is to plot y against 1/x: this gives a line through the origin with gradient 10 and recovers the same model.Test the exponential proposal on unused data
Method
Substitute x = 4 into the competing model.Reason
Agreement at the two points used to construct a model does not establish agreement with the whole data set.Working
Answer
20(1/2)⁴ = 20/16 = 1.25, but the table gives 2.5 at x = 4. The exponential proposal fails. The successive x-differences are 1, 2 and 4, so the repeated halving cannot be interpreted as a multiplier for a fixed additive step.Distinguish prediction within and beyond the data range
Method
Use the recovered model and state the range assumption.Reason
Interpolation lies within the observed x-range; extrapolation extends the model beyond it.Working
Answer
At x = 5, y = 10/5 = 2, an interpolation within 1 ≤ x ≤ 8. At x = 16 the same formula gives 10/16 = 0.625, but this is extrapolation. That value is justified only if the assumed relationship continues; the table alone does not establish it. The positive-domain model does not provide a value at x = 0.
Quick check
Compare the axes, coefficient meanings, original-scale checks and prediction assumptions separately. A different valid straight-line transformation is welcome when its equation recovers the same constants. These are self-review criteria; a straight-looking plot alone does not verify the model.
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