1.9.6 Logarithmic graphs to estimate parameters (A-level only)

What you'll learn

How logarithms turn certain non-linear relationships into straight-line equations.
Which quantities to plot for models of the form $y=ax^n$ and $y=kb^x$ .
How to estimate the parameters $a$ , $n$ , $k$ and $b$ from a line of best fit.
How to decide which model is more suitable for a set of data.

Prerequisites

Straight-line equations

A straight line can be written as

Y = m X + c,

where:

$X$ is the quantity on the horizontal axis.
$Y$ is the quantity on the vertical axis.
$m$ is the gradient, which measures the steepness of the line.
$c$ is the vertical intercept, where the line crosses the vertical axis.

For two points $X_1,Y_1)$ and $X_2,Y_2)$ on the line,

m=Y2−Y1X2−X1.m=\frac{Y_2-Y_1}{X_2-X_1}.

This familiar structure is the key to estimating parameters: you transform the variables until the model looks like $Y = m X + c$ .

Logarithm laws

You need three logarithm laws:

log⁡(uv)=log⁡u+log⁡v,log⁡(ur)=rlog⁡u,log⁡(bx)=xlog⁡b.\begin{aligned} \log(uv)&=\log u+\log v,\\ \log(u^r)&=r\log u,\\ \log(b^x)&=x\log b. \end{aligned}

In this topic, $log⁡\log$ usually means the base-10 logarithm. You could use natural logarithms, written $ln⁡\ln$ , instead, but you must use the same base throughout.

Definition

Linearisation

Linearisation is the process of transforming a non-linear relationship into a straight-line equation. It allows you to estimate unknown parameters using a graph.

Key Idea

Why logarithms help

Logarithms bring powers down in front and turn multiplication into addition. This reveals the gradient and intercept of a hidden straight line.

Power relationships

A power relationship has the form

y=ax^n,

where $a$ and $n$ are constant parameters. The parameter $n$ is the power, or index, applied to $x$ .

The graph of $y$ against $x$ is usually curved, so its gradient is not constant. Take logarithms of both sides:

log⁡y=log⁡(axn)=log⁡a+log⁡(xn)=log⁡a+nlog⁡x.\begin{aligned} \log y&=\log(ax^n)\\ &=\log a+\log(x^n)\\ &=\log a+n\log x. \end{aligned}

Reordering gives

log⁡y=nlog⁡x+log⁡a.\log y=n\log x+\log a.

Compare this with $Y = m X + c$ :

Plot $log⁡y\log y$ vertically against $log⁡x\log x$ horizontally.
The gradient is $n$ .
The vertical intercept is $log⁡a\log a$ .
Therefore, $a=10intercepta=10^{\text{intercept}}$ when base-10 logarithms are used.

Two logarithmic straight-line graphs showing how gradients and intercepts determine the parameters of power and exponential models

Example

Estimating a power model

A plot of $log⁡y\log y$ against $log⁡x\log x$ produces a line of best fit passing through the points $(0.20, 0.72)$ and $(0.80, 1.98)$ . Its vertical intercept is approximately $0.30$ . Estimate $a$ and $n$ .

Calculate the gradient of the line:
$n=1.98−0.720.80−0.20=1.260.60=2.1.n=\frac{1.98-0.72}{0.80-0.20} =\frac{1.26}{0.60} =2.1.$
The gradient represents $n$ , so $n≈2.1n\approx2.1$ .
Use the intercept $log⁡a≈0.30\log a\approx0.30$ :
$a≈100.30≈2.00.a\approx10^{0.30}\approx2.00.$
Substitute both estimates into the original model:
$y≈2.00x2.1.y\approx2.00x^{2.1}.$

Common Mistake

Forgetting the inverse logarithm

The intercept is $log⁡a\log a$ , not $a$ . If the intercept is $c$ , you must calculate $a=10^c$ for base-10 logarithms, or $a=e^c$ if natural logarithms were used.

Exponential relationships

An exponential relationship has the form

y=kb^x,

where $k$ and $b$ are constant parameters and the variable $x$ appears in the exponent.

Take logarithms:

log⁡y=log⁡(kbx)=log⁡k+log⁡(bx)=log⁡k+xlog⁡b.\begin{aligned} \log y&=\log(kb^x)\\ &=\log k+\log(b^x)\\ &=\log k+x\log b. \end{aligned}

Therefore,

log⁡y=(log⁡b)x+log⁡k.\log y=(\log b)x+\log k.

Compare this with $Y = m X + c$ :

Plot $log⁡y\log y$ vertically against $x$ horizontally.
The gradient is $log⁡b\log b$ .
The vertical intercept is $log⁡k\log k$ .
Therefore, $b=10gradientb=10^{\text{gradient}}$ and $k=10interceptk=10^{\text{intercept}}$ for base-10 logarithms.

Notice the crucial difference: for a power model you log both variables, but for an exponential model you log only $y$ .

Example

Estimating an exponential model

A plot of $log⁡y\log y$ against $x$ gives a line of best fit with gradient $0.176$ and vertical intercept $0.699$ . Estimate $k$ and $b$ .

Since the intercept is $log⁡k\log k$ ,
$k=100.699≈5.00.k=10^{0.699}\approx5.00.$
Since the gradient is $log⁡b\log b$ ,
$b=100.176≈1.50.b=10^{0.176}\approx1.50.$
Substitute the estimates into the original model:
$y≈5.00(1.50)x.y\approx5.00(1.50)^x.$

Common Mistake

Using the gradient directly

For $y=kb^x$ , the gradient is $log⁡b\log b$ , not $b$ . A gradient of $0.176$ gives $b=10^{0.176}$ , not $b = 0.176$ .

Working from a data set

A question may give measured values of $x$ and $y` rather than a ready-made logarithmic graph. You should:

Decide which model is being tested.
Calculate the required transformed values.
Plot the transformed coordinates.
Draw a straight line of best fit.
use two well-separated points on your line to find its gradient.
Read or calculate the intercept, then convert it back using an inverse logarithm.

For $y=ax^n$ , calculate $log⁡x\log x$ and $\log y`.

For $y=kb^x$ , keep $x$ unchanged and calculate only $\log y`.

Example

Transforming power-model data

Suppose two observations are $(x, y) = (2, 12.4)$ and $(5, 78.1)$ , and a power model is proposed.

Transform both coordinates using base-10 logarithms:
$(log⁡2,log⁡12.4)≈(0.301,1.093),(log⁡5,log⁡78.1)≈(0.699,1.893).\begin{aligned} (\log 2,\log 12.4)&\approx(0.301,1.093),\\ (\log 5,\log 78.1)&\approx(0.699,1.893). \end{aligned}$
Use the transformed points to estimate the gradient:
$n≈1.893−1.0930.699−0.301=0.8000.398≈2.01.n\approx\frac{1.893-1.093}{0.699-0.301} =\frac{0.800}{0.398} \approx2.01.$
Substitute one transformed point into $log⁡y=nlog⁡x+log⁡a\log y=n\log x+\log a$ :
$1.093≈2.01(0.301)+log⁡a,log⁡a≈0.488.\begin{aligned} 1.093&\approx2.01(0.301)+\log a,\\ \log a&\approx0.488. \end{aligned}$
Hence,
$a≈100.488≈3.08,a\approx10^{0.488}\approx3.08,$
giving the estimated model $y≈3.08x2.01y\approx3.08x^{2.01}$ .

Tip

Use the best-fit line

When data are scattered, calculate the gradient using two well-separated points on your line of best fit. They do not need to be original data points.

Choosing between the models

The appropriate transformation depends on the proposed relationship:

If plotting $log⁡y\log y$ against $log⁡x\log x$ gives an approximately straight line, a power model $y=ax^n$ is supported.
If plotting $log⁡y\log y$ against $x$ gives an approximately straight line, an exponential model $y=kb^x$ is supported.

A perfect line is unlikely with real data. You are looking for points that lie reasonably close to a straight line without a systematic curved pattern.

Key Idea

Straightness tests the model

The model whose required logarithmic plot is closest to a straight line is generally the more suitable model for the data.

Restrictions and interpretation

Logarithms of zero or negative numbers are not defined in real-number A-level mathematics. The basic method therefore requires the quantities being logged to be positive.

For the usual models in this topic:

a>0,k>0,b>0,a>0,\qquad k>0,\qquad b>0,

with $x > 0$ and $y > 0$ when plotting $log⁡x\log x$ against $\log y`.

If $0 < b < 1$ , then $\log b<0`, so the exponential graph has a negative gradient. This represents exponential decay.

If $n < 0$ , the log-log graph for a power model also has a negative gradient.

Common Mistake

Non-positive data

You cannot directly calculate $log⁡x\log x$ or $log⁡y\log y$ when the relevant value is zero or negative. Do not ignore such values without explaining why the proposed transformation cannot use them.

Exam technique

In the exam

Write the logarithmic equation before identifying the gradient and intercept.
Label transformed axes precisely: use $log⁡y\log y$ against $log⁡x\log x$ for $y=ax^n$ , but $log⁡y\log y$ against $x$ for $y=kb^x$ .
Use a large gradient triangle on the line of best fit and keep extra calculator figures until the final answer.
Convert logarithmic values back correctly, then state the estimated model in its original form.

Self review

Check yourself

For $y=ax^n$ , what should you plot, and what do the gradient and intercept represent?
A plot of $log⁡y\log y$ against $x$ has gradient $- 0.301$ and intercept $1.20$ . How would you estimate $b$ and $k$ ?
How could transformed graphs help you choose between a power model and an exponential model?

What you'll learn

How logarithms turn certain non-linear relationships into straight-line equations.
Which quantities to plot for models of the form $y=ax^n$ and $y=kb^x$ .
How to estimate the parameters $a$ , $n$ , $k$ and $b$ from a line of best fit.
How to decide which model is more suitable for a set of data.

Prerequisites

Straight-line equations

A straight line can be written as

Y = m X + c,

where:

$X$ is the quantity on the horizontal axis.
$Y$ is the quantity on the vertical axis.
$m$ is the gradient, which measures the steepness of the line.
$c$ is the vertical intercept, where the line crosses the vertical axis.

For two points $X_1,Y_1)$ and $X_2,Y_2)$ on the line,

m=Y2−Y1X2−X1.m=\frac{Y_2-Y_1}{X_2-X_1}.

This familiar structure is the key to estimating parameters: you transform the variables until the model looks like $Y = m X + c$ .

Logarithm laws

You need three logarithm laws:

log⁡(uv)=log⁡u+log⁡v,log⁡(ur)=rlog⁡u,log⁡(bx)=xlog⁡b.\begin{aligned} \log(uv)&=\log u+\log v,\\ \log(u^r)&=r\log u,\\ \log(b^x)&=x\log b. \end{aligned}

In this topic, $log⁡\log$ usually means the base-10 logarithm. You could use natural logarithms, written $ln⁡\ln$ , instead, but you must use the same base throughout.

Definition

Linearisation

Linearisation is the process of transforming a non-linear relationship into a straight-line equation. It allows you to estimate unknown parameters using a graph.

Key Idea

Why logarithms help

Logarithms bring powers down in front and turn multiplication into addition. This reveals the gradient and intercept of a hidden straight line.

Power relationships

A power relationship has the form

y=ax^n,

where $a$ and $n$ are constant parameters. The parameter $n$ is the power, or index, applied to $x$ .

The graph of $y$ against $x$ is usually curved, so its gradient is not constant. Take logarithms of both sides:

log⁡y=log⁡(axn)=log⁡a+log⁡(xn)=log⁡a+nlog⁡x.\begin{aligned} \log y&=\log(ax^n)\\ &=\log a+\log(x^n)\\ &=\log a+n\log x. \end{aligned}

Reordering gives

log⁡y=nlog⁡x+log⁡a.\log y=n\log x+\log a.

Compare this with $Y = m X + c$ :

Plot $log⁡y\log y$ vertically against $log⁡x\log x$ horizontally.
The gradient is $n$ .
The vertical intercept is $log⁡a\log a$ .
Therefore, $a=10intercepta=10^{\text{intercept}}$ when base-10 logarithms are used.

Two logarithmic straight-line graphs showing how gradients and intercepts determine the parameters of power and exponential models

Example

Estimating a power model

Calculate the gradient of the line:
$n=1.98−0.720.80−0.20=1.260.60=2.1.n=\frac{1.98-0.72}{0.80-0.20} =\frac{1.26}{0.60} =2.1.$
The gradient represents $n$ , so $n≈2.1n\approx2.1$ .
Use the intercept $log⁡a≈0.30\log a\approx0.30$ :
$a≈100.30≈2.00.a\approx10^{0.30}\approx2.00.$
Substitute both estimates into the original model:
$y≈2.00x2.1.y\approx2.00x^{2.1}.$

Common Mistake

Forgetting the inverse logarithm

The intercept is $log⁡a\log a$ , not $a$ . If the intercept is $c$ , you must calculate $a=10^c$ for base-10 logarithms, or $a=e^c$ if natural logarithms were used.

Exponential relationships

An exponential relationship has the form

y=kb^x,

where $k$ and $b$ are constant parameters and the variable $x$ appears in the exponent.

Take logarithms:

log⁡y=log⁡(kbx)=log⁡k+log⁡(bx)=log⁡k+xlog⁡b.\begin{aligned} \log y&=\log(kb^x)\\ &=\log k+\log(b^x)\\ &=\log k+x\log b. \end{aligned}

Therefore,

log⁡y=(log⁡b)x+log⁡k.\log y=(\log b)x+\log k.

Compare this with $Y = m X + c$ :

Plot $log⁡y\log y$ vertically against $x$ horizontally.
The gradient is $log⁡b\log b$ .
The vertical intercept is $log⁡k\log k$ .
Therefore, $b=10gradientb=10^{\text{gradient}}$ and $k=10interceptk=10^{\text{intercept}}$ for base-10 logarithms.

Notice the crucial difference: for a power model you log both variables, but for an exponential model you log only $y$ .

Example

Estimating an exponential model

A plot of $log⁡y\log y$ against $x$ gives a line of best fit with gradient $0.176$ and vertical intercept $0.699$ . Estimate $k$ and $b$ .

Since the intercept is $log⁡k\log k$ ,
$k=100.699≈5.00.k=10^{0.699}\approx5.00.$
Since the gradient is $log⁡b\log b$ ,
$b=100.176≈1.50.b=10^{0.176}\approx1.50.$
Substitute the estimates into the original model:
$y≈5.00(1.50)x.y\approx5.00(1.50)^x.$

Common Mistake

Using the gradient directly

For $y=kb^x$ , the gradient is $log⁡b\log b$ , not $b$ . A gradient of $0.176$ gives $b=10^{0.176}$ , not $b = 0.176$ .

Working from a data set

A question may give measured values of $x$ and $y` rather than a ready-made logarithmic graph. You should:

Decide which model is being tested.
Calculate the required transformed values.
Plot the transformed coordinates.
Draw a straight line of best fit.
use two well-separated points on your line to find its gradient.
Read or calculate the intercept, then convert it back using an inverse logarithm.

For $y=ax^n$ , calculate $log⁡x\log x$ and $\log y`.

For $y=kb^x$ , keep $x$ unchanged and calculate only $\log y`.

Example

Transforming power-model data

Suppose two observations are $(x, y) = (2, 12.4)$ and $(5, 78.1)$ , and a power model is proposed.

Transform both coordinates using base-10 logarithms:
$(log⁡2,log⁡12.4)≈(0.301,1.093),(log⁡5,log⁡78.1)≈(0.699,1.893).\begin{aligned} (\log 2,\log 12.4)&\approx(0.301,1.093),\\ (\log 5,\log 78.1)&\approx(0.699,1.893). \end{aligned}$
Use the transformed points to estimate the gradient:
$n≈1.893−1.0930.699−0.301=0.8000.398≈2.01.n\approx\frac{1.893-1.093}{0.699-0.301} =\frac{0.800}{0.398} \approx2.01.$
Substitute one transformed point into $log⁡y=nlog⁡x+log⁡a\log y=n\log x+\log a$ :
$1.093≈2.01(0.301)+log⁡a,log⁡a≈0.488.\begin{aligned} 1.093&\approx2.01(0.301)+\log a,\\ \log a&\approx0.488. \end{aligned}$
Hence,
$a≈100.488≈3.08,a\approx10^{0.488}\approx3.08,$
giving the estimated model $y≈3.08x2.01y\approx3.08x^{2.01}$ .

Tip

Use the best-fit line

When data are scattered, calculate the gradient using two well-separated points on your line of best fit. They do not need to be original data points.

Choosing between the models

The appropriate transformation depends on the proposed relationship:

If plotting $log⁡y\log y$ against $log⁡x\log x$ gives an approximately straight line, a power model $y=ax^n$ is supported.
If plotting $log⁡y\log y$ against $x$ gives an approximately straight line, an exponential model $y=kb^x$ is supported.

A perfect line is unlikely with real data. You are looking for points that lie reasonably close to a straight line without a systematic curved pattern.

Key Idea

Straightness tests the model

The model whose required logarithmic plot is closest to a straight line is generally the more suitable model for the data.

Restrictions and interpretation

Logarithms of zero or negative numbers are not defined in real-number A-level mathematics. The basic method therefore requires the quantities being logged to be positive.

For the usual models in this topic:

a>0,k>0,b>0,a>0,\qquad k>0,\qquad b>0,

with $x > 0$ and $y > 0$ when plotting $log⁡x\log x$ against $\log y`.

If $0 < b < 1$ , then $\log b<0`, so the exponential graph has a negative gradient. This represents exponential decay.

If $n < 0$ , the log-log graph for a power model also has a negative gradient.

Common Mistake

Non-positive data

Exam technique

In the exam

Write the logarithmic equation before identifying the gradient and intercept.
Label transformed axes precisely: use $log⁡y\log y$ against $log⁡x\log x$ for $y=ax^n$ , but $log⁡y\log y$ against $x$ for $y=kb^x$ .
Use a large gradient triangle on the line of best fit and keep extra calculator figures until the final answer.
Convert logarithmic values back correctly, then state the estimated model in its original form.

Self review

Check yourself

For $y=ax^n$ , what should you plot, and what do the gradient and intercept represent?
A plot of $log⁡y\log y$ against $x$ has gradient $- 0.301$ and intercept $1.20$ . How would you estimate $b$ and $k$ ?
How could transformed graphs help you choose between a power model and an exponential model?

What you'll learn

Prerequisites

Straight-line equations

Logarithm laws

Power relationships

Exponential relationships

Working from a data set

Choosing between the models

Restrictions and interpretation

Review 1.9.6 Logarithmic graphs to estimate parameters (A-level only) by teaching Genie

1.9.6 Logarithmic graphs to estimate parameters (A-level only)

What you'll learn

Prerequisites

Straight-line equations

Logarithm laws

Power relationships

Exponential relationships

Working from a data set

Choosing between the models

Restrictions and interpretation

Review 1.9.6 Logarithmic graphs to estimate parameters (A-level only) by teaching Genie

1.9.6 Logarithmic graphs to estimate parameters (A-level only)

What you'll learn

Prerequisites

Straight-line equations

Logarithm laws

Power relationships

Exponential relationships

Working from a data set

Choosing between the models

Restrictions and interpretation

Review 1.9.6 Logarithmic graphs to estimate parameters (A-level only) by teaching Genie

1.9.6 Logarithmic graphs to estimate parameters (A-level only) Revision Guide

1.9.6 Logarithmic graphs to estimate parameters (A-level only)

What you'll learn

Prerequisites

Straight-line equations

Logarithm laws

Power relationships

Exponential relationships

Working from a data set

Choosing between the models

Restrictions and interpretation

Review 1.9.6 Logarithmic graphs to estimate parameters (A-level only) by teaching Genie

1.9.6 Logarithmic graphs to estimate parameters (A-level only) Revision Guide