Scaling using logarithms and linearising data AI HL

IB Mathematics: Applications and Interpretation HL · Unit 2: Functions

Lesson 2.9 · Scaling Using Logarithms and Linearising Data

Manage large ranges, uncover patterns and use straight-line fits to build models.

Faith connection

Paul encourages believers to “Be kindly affectioned one to another with brotherly love” (Romans 12:10, KJV). Loving others means noticing them and treating them with care. In mathematics, changing a graph’s scale can reveal patterns that were difficult to see. As you learn to look closely at data, ask Jesus, our Savior and Lord, to help you look closely at the needs of people around you and respond with love.

Learning objectives

By the end of this lesson, you should be able to:

explain why logarithmic scales are useful for very large ranges of positive values;
interpret semi-logarithmic and log-log graphs;
linearise exponential and power models using logarithms;
recover model parameters from the gradient and intercept of a best-fit line;
check predictions in the original units and assess whether a model is reasonable.

1. Why use a logarithmic scale?

On an ordinary axis, equal distances represent equal differences. On a logarithmic axis, equal distances represent equal ratios. For example, $10$ , $100$ , $1000$ and $10\,000$ are equally spaced on a base-10 logarithmic scale because each value is ten times the previous one.

A logarithmic scale can make values that differ by many orders of magnitude easier to display and compare. Only positive values can be placed on a standard logarithmic axis: $\log0$ is undefined, as is the real logarithm of a negative number.

Worked example · A wide range of populations.

Suppose four samples contain $10$ , $100$ , $1000$ and $10\,000$ bacteria. On an ordinary vertical axis reaching $10\,000$ , the first two points appear close to zero. On a base-10 logarithmic axis, their coordinates are:

$\log_{10}10=1,\quad \log_{10}100=2,\quad \log_{10}1000=3,\quad \log_{10}10\,000=4$

The four values appear at equally spaced positions. This spacing represents repeated multiplication by $10$ , not repeated addition by a fixed number.

2. Semi-logarithmic graphs

A semi-logarithmic graph uses an ordinary scale on one axis and a logarithmic scale on the other. For exponential data, we often keep $x$ on an ordinary scale and use $\ln y$ or $\log_{10}y$ on the vertical axis.

If $y=Ae^{kx}$ , where $A>0$ , taking natural logarithms gives a straight-line equation:

$\ln y=\ln A+kx$

Thus, on a plot of $\ln y$ against $x$ , the gradient is $k$ and the vertical intercept is $\ln A$ . Recover the original coefficient using $A=e^{\text{intercept}}$ .

Worked example · Exponential growth.

Suppose a best-fit line for a plot of $\ln y$ against $x$ is:

$\ln y=1.61+0.30x$

Exponentiate to return to the original variables:

$y=e^{1.61}e^{0.30x}\approx5.00e^{0.30x}$

The continuous growth parameter is $0.30$ per unit of $x$ . The multiplier for each increase of one unit in $x$ is $e^{0.30}\approx1.35$ , corresponding to approximately $35\%$ growth per unit.

Another common form: If $y=Ab^x$ , then $\ln y=\ln A+x\ln b$ . The straight-line gradient is $\ln b$ , so $b=e^{\text{gradient}}$ .

3. Linearising exponential data

To test an exponential model from positive measured outputs, calculate the logarithm of each output and plot those transformed values against the original inputs. If the transformed points lie approximately along a straight line, an exponential model may be reasonable.

Worked example · Find a model from two points.

A quantity is $20$ at $x=0$ and $80$ at $x=4$ . Assume $y=Ae^{kx}$ . The first point gives $A=20$ . The second gives:

$80=20e^{4k}\quad\Longrightarrow\quad e^{4k}=4$

$k=\frac{\ln4}{4}\approx0.347$

Therefore:

$y=20e^{(\ln4)x/4}$

At $x=2$ , the model gives $20e^{(\ln4)/2}=40$ . More measured points would help determine whether an exponential pattern actually fits the situation.

4. Log-log graphs and power relationships

A log-log graph uses logarithmic scales on both axes. It is useful for investigating a power model:

$y=Ax^n,\qquad x>0,\ y>0,\ A>0$

Taking natural logarithms gives:

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

On a plot of $\ln y$ against $\ln x$ , the gradient is the power $n$ , and the vertical intercept is $\ln A$ . Therefore $A=e^{\text{intercept}}$ .

Worked example · A power law.

A best-fit line on a plot of $\ln y$ against $\ln x$ is:

$\ln y=1.10+1.50\ln x$

The corresponding power model is:

$y=e^{1.10}x^{1.50}\approx3.00x^{1.50}$

The gradient $1.50$ is an exponent, not an ordinary rate of increase. For example, doubling $x$ multiplies the predicted output by $2^{1.50}\approx2.83$ .

5. Calculate power-model parameters from data

When two positive data points are assumed to follow a power model exactly, divide their equations to remove the coefficient $A$ .

Worked example.

Suppose $y=Ax^n$ , with $y=12$ at $x=2$ and $y=96$ at $x=8$ . Divide:

$\frac{96}{12}=\left(\frac82\right)^n$

$8=4^n\quad\Longrightarrow\quad n=\frac{\ln8}{\ln4}=1.5$

Use $(2,12)$ to obtain the coefficient:

$A=\frac{12}{2^{1.5}}=3\sqrt2\approx4.24$

$y=3\sqrt2\,x^{1.5}$

Check using $x=8$ : the model gives $3\sqrt2(8^{1.5})=96$ .

6. Use a best-fit line to estimate parameters

Real measurements rarely lie exactly on a line after transformation. A calculator can find a straight-line fit for the transformed data. Make sure the line is fitted to the variables you actually plotted.

Suspected original model	Plot a straight line using	If its line is $Y=mX+c$
$y=Ae^{kx}$	$X=x$ , $Y=\ln y$	$k=m$ and $A=e^c$
$y=Ab^x$	$X=x$ , $Y=\ln y$	$b=e^m$ and $A=e^c$
$y=Ax^n$	$X=\ln x$ , $Y=\ln y$	$n=m$ and $A=e^c$

Worked example · Approximate exponential fit.

The measured outputs at $x=0,1,2,3$ are $50,73,108,158$ . A linear fit to the points $(x,\ln y)$ gives approximately:

$\ln y\approx3.910+0.384x$

Undo the logarithm:

$y\approx e^{3.910}e^{0.384x}\approx49.9e^{0.384x}$

The model predicts approximately $341$ at $x=5$ . This is an extrapolation because the measured inputs end at $x=3$ .

7. Interpret the shape, not just the line

A straight line on a logarithmic plot has a different meaning from a straight line on an ordinary plot:

A straight line on ordinary axes suggests a linear relationship.
A straight line on a semi-log plot with a logarithmic vertical axis suggests an exponential relationship.
A straight line on a log-log plot suggests a power relationship.

These interpretations depend on the axes and on the data being appropriate for the model. You do not need to draw or sketch these special graphs in this topic, but you should be able to interpret them.

Worked example · Identify the model.

A graph of $\log_{10}y$ against $x$ is approximately a straight line with gradient $-0.20$ . Since $\log_{10}y=\log_{10}A+x\log_{10}b$ , we have:

$\log_{10}b=-0.20\quad\Longrightarrow\quad b=10^{-0.20}\approx0.631$

The model predicts that $y$ is multiplied by about $0.631$ for each unit increase in $x$ , equivalent to a decrease of about $36.9\%$ per unit.

8. Check whether linearisation is appropriate

Taking logarithms requires positive values. A shifted exponential such as $y=Ae^{kx}+d$ does not generally produce a straight line when you plot $\ln y$ against $x$ . If $d$ is known and $y-d>0$ , then plot $\ln(y-d)$ against $x$ instead.

Worked example · A nonzero background level.

A sensor reading is modelled by $R(t)=5+20e^{-0.4t}$ . The reading approaches $5$ , not zero. Subtract the background level first:

$R(t)-5=20e^{-0.4t}$

$\ln(R(t)-5)=\ln20-0.4t$

A plot of $\ln(R-5)$ against $t$ is linear with gradient $-0.4$ . A plot of $\ln R$ against $t$ is generally curved.

Calculator workflow

Enter the original positive input and output values in lists.
For an exponential investigation, make a new list containing $\ln y$ and fit a line against $x$ .
For a power investigation, make lists containing both $\ln x$ and $\ln y$ , then fit a line to those lists.
Record the gradient and intercept with enough digits before converting them back to the original parameters.
Graph the resulting model with the original data. Interpret predictions in the original units.

If a calculator reports a strong correlation for transformed points, also inspect whether the original model makes sense in the context. A straight-looking transformed plot does not justify unlimited extrapolation.

IB-style Paper 1 · Skill practice

Question 1 · Understand a logarithmic axis

On a base-10 logarithmic vertical axis, the marks $10$ , $100$ , $1000$ and $10\,000$ are equally spaced. Explain why. Can a value of $0$ be plotted on this axis?

Show solution

Each successive value is multiplied by $10$ , and their base-10 logarithms are $1,2,3,4$ . Zero cannot be plotted because $\log_{10}0$ is undefined.

Question 2 · Exponential line

A straight-line fit to $\ln y$ against $x$ gives $\ln y=2.30+0.18x$ . Find an exponential model for $y$ and its predicted value at $x=4$ . Give numerical answers to three significant figures.

Show solution

$y=e^{2.30}e^{0.18x}\approx9.97e^{0.18x}$ . At $x=4$ , $y=e^{2.30+0.72}=e^{3.02}\approx20.5$ .

Question 3 · Power line

A straight-line fit to $\ln y$ against $\ln x$ gives $\ln y=0.70+2.40\ln x$ . Write the power model and find the factor by which $y$ changes when $x$ doubles.

Show solution

$y=e^{0.70}x^{2.40}\approx2.01x^{2.40}$ . Doubling $x$ multiplies $y$ by $2^{2.40}\approx5.28$ .

Question 4 · Identify the axes

Data form an approximately straight line when $\ln y$ is plotted against $\ln x$ , but a curved pattern when $\ln y$ is plotted against $x$ . Which model family is better supported: exponential or power? Explain.

Show solution

A power model $y=Ax^n$ is better supported, because $\ln y=\ln A+n\ln x$ is linear in $\ln x$ . An exponential model $y=Ae^{kx}$ would instead make $\ln y$ linear in $x$ .

Question 5 · Base-10 logs

On a semi-log plot, a fitted line has equation $\log_{10}y=1.5-0.30x$ . Write the corresponding model $y=Ab^x$ . Give $A$ and $b$ to three significant figures.

Show solution

$y=10^{1.5}(10^{-0.30})^x$

Thus $A=10^{1.5}\approx31.6$ and $b=10^{-0.30}\approx0.501$ . An appropriate rounded model is $y\approx31.6(0.501)^x$ .

Question 6 · Shifted exponential

A model is $f(t)=8+30e^{-0.25t}$ . State which quantity should be logarithmically transformed to obtain a straight-line relationship with $t$ , and give the gradient of that line.

Show solution

Subtract the background value $8$ and plot $\ln(f(t)-8)$ against $t$ . Since $\ln(f(t)-8)=\ln30-0.25t$ , the gradient is $-0.25$ .

Applications and interpretation

Application 1 · Bacterial growth

A culture begins with $500$ bacteria. Measurements suggest that a plot of $\ln N$ against time $t$ , in hours, has gradient $0.28$ .

Construct a model $N(t)$ .
Find the multiplier for each additional hour.
Estimate when the model predicts $2000$ bacteria.

Show solution

The initial value gives $N(t)=500e^{0.28t}$ . The hourly multiplier is $e^{0.28}\approx1.32$ . Solve $500e^{0.28t}=2000$ :

$t=\frac{\ln4}{0.28}\approx4.95\text{ hours}$

This prediction assumes that the early growth conditions remain suitable.

Application 2 · Scaling a physical measurement

An experiment suggests $F=Av^n$ , where $v$ is speed in m/s and $F$ is force in newtons. Measurements give $F=18$ N at $v=3$ m/s and $F=72$ N at $v=6$ m/s.

Find $n$ and $A$ .
Predict the force at $v=10$ m/s.
Interpret the exponent.

Show solution

Dividing the observations gives $72/18=(6/3)^n$ , so $4=2^n$ and $n=2$ . Then $18=A(3)^2$ , giving $A=2$ in the corresponding units. The model is $F=2v^2$ . At $10$ m/s it predicts $200$ N. The exponent means that doubling speed multiplies the force by $2^2=4$ , within the model’s applicable range.

Application 3 · Website traffic

The daily visits to a new website are recorded as follows:

Week $t$	0	1	2	3
Visits per day	50	73	108	158

A fitted line for $\ln V$ against $t$ is approximately $\ln V=3.910+0.384t$ .

Recover the exponential model $V(t)$ .
Estimate visits per day at week $5$ .
Explain why the week $5$ prediction needs caution.

Show solution

$V(t)=e^{3.910}e^{0.384t}\approx49.9e^{0.384t}$ . At week $5$ , the model predicts approximately $341$ daily visits. The supplied measurements end at week $3$ , so week $5$ is an extrapolation; traffic growth might slow or change.

Application 4 · Compare two interpretations

Two reports claim that an environmental measurement “increases by the same amount each year.” The first report shows a straight line on ordinary axes. The second shows a straight line only when the vertical axis is logarithmic.

Explain the likely difference between the two patterns.
Which pattern could have a constant percentage increase?
Why is identifying the axis scale important before describing a trend?

Show solution

The first report suggests approximately constant absolute increases, as in a linear model. The second suggests approximately constant multiplicative changes, as in an exponential model. The second pattern could have a constant percentage increase. Equal vertical distances mean equal differences on ordinary axes but equal ratios on logarithmic axes, so the same-looking straight line can express a different relationship.

Paper 1 checklist

Name the variables on each axis. Use logarithms only for positive values. For a semi-log fit, convert the line’s intercept back to the initial multiplier and its gradient back to a growth or decay factor. For a log-log fit, interpret the gradient as an exponent. Return predictions to the original units and assess extrapolation.

Unit 2 reflection · Loving others

Throughout this unit, functions helped us examine relationships, changes and patterns. Our relationships with people also call for careful attention. Scripture asks us to consider whether our words and actions show love toward our brothers and sisters:

1 John 2:9–11

9 Anyone who claims to be in the light but hates a brother or sister is still in the darkness. 10 Anyone who loves their brother and sister lives in the light, and there is nothing in them to make them stumble. 11 But anyone who hates a brother or sister is in the darkness and walks around in the darkness. They do not know where they are going, because the darkness has blinded them.

Think of one person you can encourage, forgive or help this week. Bring your heart to Jesus, the Savior and Lord, and ask him to lead you in his light as you love others.

Lesson 2.9 · Scaling Using Logarithms and Linearising Data

Faith connection

Learning objectives

1. Why use a logarithmic scale?

2. Semi-logarithmic graphs

3. Linearising exponential data

4. Log-log graphs and power relationships

5. Calculate power-model parameters from data

6. Use a best-fit line to estimate parameters

7. Interpret the shape, not just the line

8. Check whether linearisation is appropriate

Calculator workflow

IB-style Paper 1 · Skill practice

Question 1 · Understand a logarithmic axis

Question 2 · Exponential line

Question 3 · Power line

Question 4 · Identify the axes

Question 5 · Base-10 logs

Question 6 · Shifted exponential

Applications and interpretation

Application 1 · Bacterial growth

Application 2 · Scaling a physical measurement

Application 3 · Website traffic

Application 4 · Compare two interpretations

Paper 1 checklist

Unit 2 reflection · Loving others

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IB Mathematics: Applications and Interpretation HL 6 Sections80 LessonsLifetime

Curriculum

Scaling using logarithms and linearising data AI HL

Faith connection

Learning objectives

1. Why use a logarithmic scale?

2. Semi-logarithmic graphs

3. Linearising exponential data

4. Log-log graphs and power relationships

5. Calculate power-model parameters from data

6. Use a best-fit line to estimate parameters

7. Interpret the shape, not just the line

8. Check whether linearisation is appropriate

Calculator workflow

IB-style Paper 1 · Skill practice

Question 1 · Understand a logarithmic axis

Question 2 · Exponential line

Question 3 · Power line

Question 4 · Identify the axes

Question 5 · Base-10 logs

Question 6 · Shifted exponential

Applications and interpretation

Application 1 · Bacterial growth

Application 2 · Scaling a physical measurement

Application 3 · Website traffic

Application 4 · Compare two interpretations

Paper 1 checklist

Unit 2 reflection · Loving others

Related

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