Logarithmic modelling | Year 12

01 / LEARN

Understand the idea.

Logarithmic response and diminishing increases

For $R (t) = a + b ln (1 + \frac{t}{c})$ with $b, c > 0$ and $t \geq 0$ , the response increases while its rate $R^{'} (t) = \frac{b}{c + t}$ decreases. The logarithm input is dimensionless because $t$ and $c$ use the same time units.

Substitute observed data to fit unknown constants before predicting a new value. Two fitted points establish parameter values for the assumed formula, but do not prove the mechanism or justify behaviour far outside the observed interval.

Multiplicative comparisons become differences

On a scale $S = k {log}_{10} (\frac{I}{I_{0}})$ , positive $I$ and $I_{0}$ form a dimensionless ratio. Subtracting scale values combines their logarithms, so the difference determines an intensity ratio after exponentiating.

Keep ratios separate from differences and percentages. State which quantity is larger and by what factor. Also check whether a predicted time or response lies within a credible model range; indefinite logarithmic growth may ignore a real upper limit.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Fit a learning model

A training score is modelled by $S (t) = a + b ln (t + 1)$ for $t \geq 0$ days. It is $40$ at day $0$ and $70$ at day $3$ . Find $a, b$ and the predicted day-8 score.

Worked solution

1
Use the initial score to find a.
$S (0) = a + b ln 1 = a = 40$
The logarithm term vanishes at day zero.
2
Use the day-3 score to find b.
$70 = 40 + b ln 4 ⟹ b = \frac{30}{ln 4}$
The observed increase of 30 determines the remaining coefficient.
3
Substitute day 8 into the fitted model.
$S (8) = 40 + \frac{30}{ln 4} ln 9 = 40 + \frac{30 ln 9}{ln 4}$
The logarithm input at day 8 is 8+1=9; exact constants avoid premature rounding.
4
Calculate and interpret the prediction.
$S (8) = 87.548875 \dots \approx 87.5$
This is the predicted score from the fitted model; day 8 extends beyond the two supplied observations.

Answer

$a = 40$ , $b = 30 / ln 4$ ; day-8 score approximately $87.5$ .

EXAMPLE 02

Compare a logarithmic scale

A supplied sound model is $L = 10 {log}_{10} (I / I_{0})$ . Compare the intensities at $L = 60$ and $L = 45$ .

Worked solution

1
Write the two scale equations.
$60 = 10 {log}_{10} \frac{I_{60}}{I_{0}}, 45 = 10 {log}_{10} \frac{I_{45}}{I_{0}}$
Both intensities are compared with the same positive reference.
2
Subtract and combine logarithms.
$15 = 10 {log}_{10} (\frac{I_{60}}{I_{45}})$
The quotient law cancels the common reference intensity.
3
Exponentiate to find the intensity ratio.
${log}_{10} \frac{I_{60}}{I_{45}} = 1.5 ⟹ \frac{I_{60}}{I_{45}} = 10^{1.5} \approx 31.6$
Base-10 exponentiation reverses the logarithm.
4
State the comparison.
$I_{60} \approx 31.6 I_{45}$
The 60-unit level has about 31.6 times the intensity, rather than an intensity difference of 15.

Answer

The 60-unit level has approximately $31.6$ times the intensity.

03 / YOUR TURN

Try it yourself.

Check your understanding

For $C (t) = 20 + 8 ln (t + 1)$ , where $t \geq 0$ is measured in days, find the instantaneous change at $t = 3$ .

Show answer & reasoning

$2$ score units per day.

Differentiate the composite logarithm: $C^{'} (t) = \frac{8}{t + 1}$ . At $t = 3$ days this gives $\frac{8}{4} = 2$ score units per day, the instantaneous rate rather than the accumulated score increase.

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6 worksheet questions · 17 marks

Open worksheet Worked solutions

Read the questions online

For $P (t) = 15 + 6 ln (t + 1)$ , $t \geq 0$ weeks, find $P (0)$ and $P (3)$ .
3 marks
Find when the model $R (t) = 10 + 5 ln (t + 1)$ first reaches $20$ , with $t$ in hours.
3 marks
For $R (t) = 10 + 5 ln (t + 1)$ with $t \geq 0$ measured in hours, find and interpret $R^{'} (4)$ .
3 marks
A scale is $S = 20 {log}_{10} (\frac{A}{A_{0}})$ . What amplitude ratio gives a rise of 12 units?
3 marks
The model $F (x) = a + b ln x$ gives $F (1) = 8$ and $F (e^{2}) = 18$ . Find $a, b$ .
3 marks
A score $30 + 12 ln (t + 1)$ is fitted using data from days 0–10. Explain one reason not to trust its prediction for day 1000.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

For $P (t) = 15 + 6 ln (t + 1)$ , $t \geq 0$ weeks, find $P (0)$ and $P (3)$ .
3 marks
Substitute the times; $ln 1 = 0$ .
$15$ and $15 + 6 ln 4 \approx 23.3$
Find when the model $R (t) = 10 + 5 ln (t + 1)$ first reaches $20$ , with $t$ in hours.
3 marks
$ln (t + 1) = 2$ , then exponentiate and subtract 1.
$t = e^{2} - 1 \approx 6.39$ hours
For $R (t) = 10 + 5 ln (t + 1)$ with $t \geq 0$ measured in hours, find and interpret $R^{'} (4)$ .
3 marks
$R^{'} (t) = \frac{5}{t + 1}$ , then substitute $t = 4$ .
$1$ response unit per hour.
A scale is $S = 20 {log}_{10} (\frac{A}{A_{0}})$ . What amplitude ratio gives a rise of 12 units?
3 marks
$12 = 20 {log}_{10} (\frac{A_{2}}{A_{1}})$ . Exponentiate base 10.
$10^{0.6} \approx 3.98$
The model $F (x) = a + b ln x$ gives $F (1) = 8$ and $F (e^{2}) = 18$ . Find $a, b$ .
3 marks
$ln 1 = 0$ gives $a = 8$ . Then $18 = 8 + 2 b$ .
$a = 8$ , $b = 5$
A score $30 + 12 ln (t + 1)$ is fitted using data from days 0–10. Explain one reason not to trust its prediction for day 1000.
2 marks
The formula grows without bound. A real learning score may have a maximum or change its pattern.
The prediction extrapolates far beyond the measured interval and may ignore a real score limit.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: logarithmic functions and their applications; reciprocal-linear antiderivatives are taught in the Integration chapter.

Official curriculum source ↗

In this lesson

Understand the idea.

Logarithmic response and diminishing increases

Multiplicative comparisons become differences

See the method in action.

Fit a learning model

Use the initial score to find a.

Use the day-3 score to find b.

Substitute day 8 into the fitted model.

Calculate and interpret the prediction.

Compare a logarithmic scale

Write the two scale equations.

Subtract and combine logarithms.

Exponentiate to find the intensity ratio.

State the comparison.

Try it yourself.

Check your understanding

Keep practising