Linearising exponential and power data

Linearising data means transforming variables so that a proposed nonlinear relationship becomes a straight-line relationship. A straight line is easier to recognise and analyse because its gradient and intercept reveal the parameters of the original model.

The two central results are

$$\boxed{y=Ab^x\quad\Longrightarrow\quad \ln y=x\ln b+\ln A}$$

and

$$\boxed{y=Ax^n\quad\Longrightarrow\quad \ln y=n\ln x+\ln A.}$$

They look similar, but require different graphs. For an exponential model, plot $\ln y$ against $x$. For a power model, plot $\ln y$ against $\ln x$.

Which graph should you plot?

Proposed model	Linear form	Horizontal axis	Vertical axis	Gradient	Intercept
$y=mx+c$	$y=mx+c$	$x$	$y$	$m$	$c$
$y=Ab^x$	$\ln y=(\ln b)x+\ln A$	$x$	$\ln y$	$\ln b$	$\ln A$
$y=Ae^{kx}$	$\ln y=kx+\ln A$	$x$	$\ln y$	$k$	$\ln A$
$y=Ax^n$	$\ln y=n\ln x+\ln A$	$\ln x$	$\ln y$	$n$	$\ln A$

The phrase plot $P$ against $Q$ means put $P$ on the vertical axis and $Q$ on the horizontal axis.

The same four observations appear on both graphs. Transforming the vertical coordinate changes the curve into a line; it does not change the original measurements.

Before you begin

You should be able to:

• use exponential functions;
• apply the laws of logarithms;
• interpret the gradient and intercept in the equation of a straight line;
• use a calculator to find logarithms and perform linear regression.

Why linearisation works

A straight line has the form

$$Y=mX+c,$$

where $X$ is the horizontal variable, $Y$ is the vertical variable, $m$ is the gradient and $c$ is the vertical intercept.

Linearisation does not claim that the original graph of $y$ against $x$ is straight. It defines transformed variables $X$ and $Y$ so that the transformed equation has the form $Y=mX+c$.

The comparison with $Y=mX+c$ should always be explicit. It prevents the most common error: reading the correct gradient from the wrong axes.

Exponential models

Suppose

$$y=Ab^x,$$

where $A>0$, $b>0$ and $b\ne1$. Taking natural logarithms gives

$$\begin{aligned} \ln y&=\ln(Ab^x)\\ &=\ln A+\ln(b^x)\\ &=\ln A+x\ln b. \end{aligned}$$

Rearrange this as

$$\underbrace{\ln y}_{Y} =\underbrace{\ln b}_{m}\underbrace{x}_{X} +\underbrace{\ln A}_{c}.$$

Therefore a plot of $\ln y$ on the vertical axis against $x$ on the horizontal axis should be a straight line with

$$m=\ln b, \qquad c=\ln A.$$

Recover the original parameters by undoing the logarithms:

$$\boxed{b=e^m,\qquad A=e^c.}$$

If the original model is written as $y=Ae^{kx}$, then

$$\ln y=kx+\ln A,$$

so the gradient is directly $k$.

Worked example 1: transform an equation

The variables $x$ and $y$ are believed to satisfy

$$y=7(1.4)^x.$$

Find the equation of the straight line obtained by plotting $\ln y$ against $x$.

Take natural logarithms:

$$\ln y=\ln 7+x\ln 1.4.$$

Hence the straight-line equation is

$$\ln y=0.33647\ldots x+1.94591\ldots.$$

The gradient is positive because $1.4>1$, so the original model represents growth.

Worked example 2: recover an exponential model

A plot of $\ln y$ against $x$ produces the line

$$\ln y=1.7-0.24x.$$

Write the relationship between $x$ and $y$ in the forms $y=Ae^{kx}$ and $y=Ab^x$.

Exponentiate both sides:

$$\begin{aligned} y&=e^{1.7-0.24x}\\ &=e^{1.7}e^{-0.24x}. \end{aligned}$$

Therefore

$$y=5.4739\ldots e^{-0.24x},$$

so, to three significant figures,

$$\boxed{y=5.47e^{-0.24x}.}$$

For the form $y=Ab^x$,

$$b=e^{-0.24}=0.7866\ldots,$$

giving

$$\boxed{y=5.47(0.787)^x.}$$

The negative gradient means $0<b<1$, which is consistent with exponential decay.

Worked example 3: linearise a table of data

The following measurements are believed to follow $y=Ae^{kx}$.

$x$	$0$	$2$	$4$	$6$
$y$	$3.00$	$4.48$	$6.68$	$9.96$

Add a row of transformed values, keeping enough decimal places for the calculation:

$x$	$0$	$2$	$4$	$6$
$\ln y$	$1.099$	$1.499$	$1.899$	$2.299$

The increments in $\ln y$ are approximately constant for equal increments in $x$, so a plot of $\ln y$ against $x$ is approximately linear.

Using the first and last transformed points,

$$m\approx\frac{2.299-1.099}{6-0}=0.200.$$

The intercept is approximately $1.099$, so

$$k\approx0.200, \qquad A\approx e^{1.099}=3.00.$$

The model is therefore

$$\boxed{y\approx3.00e^{0.200x}.}$$

With real data, use all the transformed points in a linear regression rather than estimating from two arbitrarily chosen data points.

Power models

Suppose

$$y=Ax^n,$$

where $A>0$ and, for logarithms, $x>0$ and $y>0$. Taking natural logarithms gives

$$\begin{aligned} \ln y&=\ln(Ax^n)\\ &=\ln A+n\ln x. \end{aligned}$$

Now compare with $Y=mX+c$:

$$\underbrace{\ln y}_{Y} =\underbrace{n}_{m}\underbrace{\ln x}_{X} +\underbrace{\ln A}_{c}.$$

Therefore a plot of $\ln y$ against $\ln x$ should be a straight line with

$$\boxed{n=m,\qquad A=e^c.}$$

The gradient is the power itself. This is the key distinction from an exponential model.

Worked example 4: recover a power model

A plot of $\ln y$ against $\ln x$ has gradient $1.6$ and vertical intercept $-0.7$. Find the model $y=Ax^n$.

Since the gradient equals $n$,

$$n=1.6.$$

Since the intercept equals $\ln A$,

$$\ln A=-0.7$$

and therefore

$$A=e^{-0.7}=0.496585\ldots.$$

Thus

$$\boxed{y\approx0.497x^{1.6}.}$$

Do not set $A=-0.7$. The intercept is $\ln A$, not $A$.

Worked example 5: use two transformed points

Variables $x$ and $y$ satisfy $y=Ax^n$. On a graph of $\log_{10}y$ against $\log_{10}x$, the straight line passes through $(0,0.4)$ and $(2,3.0)$. Find $A$ and $n$.

The gradient is

$$n=\frac{3.0-0.4}{2-0}=1.3.$$

The intercept is

$$\log_{10}A=0.4,$$

so

$$A=10^{0.4}=2.5118\ldots.$$

Therefore

$$\boxed{y\approx2.51x^{1.3}.}$$

This example uses base $10$ rather than base $e$. The method is unchanged, but the inverse operation must match the logarithm used: $A=10^c$ for common logarithms and $A=e^c$ for natural logarithms.

Worked example 6: build and use a model

For a particular material, breaking load $F$ is modelled by $F=Ad^n$, where $d$ is diameter. A log-log regression gives

$$\ln F=2.08\ln d+3.15.$$

Find the model and estimate $F$ when $d=4.5$.

Comparing with

$$\ln F=n\ln d+\ln A,$$

gives

$$n=2.08, \qquad A=e^{3.15}=23.336\ldots.$$

Hence

$$F\approx23.336d^{2.08}.$$

At $d=4.5$,

$$F\approx23.336(4.5)^{2.08}=533.7\ldots.$$

Therefore

$$\boxed{F\approx534}$$

in the stated units. Use the unrounded values of $A$ and $n$ in the final calculation if they are available.

A useful diagnostic

If both an exponential and a power model seem plausible:

1. plot $\ln y$ against $x$ to test an exponential model;
2. plot $\ln y$ against $\ln x$ to test a power model;
3. compare the patterns of residuals and the strength of the linear fits;
4. use the context and mechanism, not correlation alone, to select a model.

A high correlation coefficient for transformed data supports linearity over the observed range. It does not prove that the proposed law is true for all values.

Using regression carefully

For transformed data, enter the appropriate transformed variables into two calculator lists and perform linear regression in the form

$$Y=a+bX.$$

Calculator letters vary, so translate the output by meaning:

• $b$ is the gradient;
• $a$ is the vertical intercept;
• the original parameters follow from the comparison table above.

Record more digits than the final answer requires. Premature rounding of logarithms, gradients or intercepts can noticeably change the recovered model.

If a question asks you to draw a line of best fit by hand, choose two well-separated points on the line, not necessarily two measured data points. Then calculate

$$m=\frac{Y_2-Y_1}{X_2-X_1}.$$

Widely separated points reduce the effect of reading error.

Domain, units and limitations

Logarithms require positive inputs

To calculate $\ln x$ and $\ln y$ as real numbers, both $x$ and $y$ must be positive. Zero and negative observations cannot simply be included in a logarithmic transformation.

This does not automatically mean the underlying situation is impossible. It means that this particular transformation and model need reconsideration. A shifted model such as $y=C+Ab^x$ does not become linear merely by plotting $\ln y$ against $x$ because

$$\ln(C+Ab^x)\ne \ln C+\ln(Ab^x).$$

Logs and dimensional quantities

Strictly, a logarithm acts on a dimensionless ratio. In a data analysis, writing $\ln x$ is shorthand for logging the numerical value of $x$ in fixed stated units, or for $\ln(x/x_0)$ using a reference value $x_0$.

Changing units can alter the intercept, and sometimes the numerical value of $A$, but it does not alter the essential fitted relationship when conversions are handled consistently. Always state the units of the original variables and use one unit system throughout.

Transformation changes the error structure

Least-squares regression on $\ln y$ minimises vertical discrepancies in $\ln y$, not in $y$. It is therefore especially natural when errors are proportional or multiplicative. A good transformed fit should still be checked against the original data and context.

Interpolation and extrapolation

An estimate within the observed range is an interpolation. An estimate beyond it is an extrapolation and is less reliable. Linear appearance after transformation does not justify extreme predictions.

Common misconceptions

”Taking logs makes any curve straight”

No. It makes particular model families straight. If the proposed model is unsuitable, the transformed points will still show curvature or another systematic pattern.

”The gradient of $\ln y$ against $x$ is the base”

For $y=Ab^x$, the gradient is $\ln b$, so $b=e^m$. The gradient is directly the exponent coefficient only when the model is written $y=Ae^{kx}$.

”A power model needs only the $y$ values logged”

For $y=Ax^n$, both variables must be logged:

$$\ln y=n\ln x+\ln A.$$

Plotting $\ln y$ against $x$ tests an exponential model instead.

”The intercept is $A$”

For both principal transformations, the intercept is $\ln A$. Exponentiate it to recover $A$.

”The base of the logarithm changes the power”

It does not. For a power model, the gradient is $n$ with any consistent logarithm base. Only the interpretation of the intercept changes from $A=e^c$ to $A=10^c$, for example.

Self-check

1. Identify the axes

Which graph should be linear if $y=6(0.8)^x$?

Answer

Plot $\ln y$ vertically against $x$ horizontally. The line has gradient $\ln 0.8<0$ and intercept $\ln 6$.

2. Recover an exponential model

A plot of $\ln y$ against $x$ has gradient $0.35$ and intercept $1.2$. Find the model in the form $y=Ae^{kx}$.

Answer

$$k=0.35, \qquad A=e^{1.2}=3.3201\ldots.$$

Therefore

$$y\approx3.32e^{0.35x}.$$

3. Recover a power model

A plot of $\log_{10}y$ against $\log_{10}x$ has equation

$$\log_{10}y=2.4\log_{10}x-0.3.$$

Find $A$ and $n$ in $y=Ax^n$.

Answer

The gradient gives $n=2.4$. The intercept gives

$$\log_{10}A=-0.3,$$

so

$$A=10^{-0.3}=0.5012\ldots.$$

Thus $y\approx0.501x^{2.4}$.

4. Spot the error

A student linearises $y=Ax^n$ and writes $\ln y=n\ln x+A$. What has gone wrong?

Answer

They have treated $\ln A$ as $A$. The correct transformation is

$$\ln y=n\ln x+\ln A.$$

5. Interpret the evidence

The correlation coefficient between $x$ and $\ln y$ is $-0.998$. What does this suggest, and what does it not prove?

Answer

It suggests that an exponential decay model may fit the observed range very well because $\ln y$ is strongly and negatively linearly related to $x$. It does not prove causation, establish the mechanism, or guarantee reliable extrapolation outside the data range.

Exam checklist

Before finishing a linearisation question, check that you have:

• taken logarithms of the complete model;
• expanded them using the log laws correctly;
• identified the transformed $X$ and $Y$ variables;
• compared the equation explicitly with $Y=mX+c$;
• interpreted the gradient and intercept for the stated model;
• undone the logarithm using the matching base;
• kept unrounded calculator values until the final answer;
• stated the final model in the requested form;
• commented on fit and extrapolation only as strongly as the evidence allows.

Next steps

• Use these models in exponential growth and decay.
• Practise solving for unknown inputs in exponential and logarithmic equations.
• Develop model selection and validation in functions in modelling.
• Review how to interpret conclusions in context in interpreting context.