Logarithmic modelling | Year 12

01 / LEARN

The idea, step by step

Logarithmic response and diminishing increases

A model such as $R (t) = a + b ln (1 + t / c)$ , with $c > 0$ , can describe a response whose increases get smaller. Its logarithm input must be dimensionless; $t / c$ compares quantities with the same units. The derivative is $R^{'} (t) = b / (c + t)$ , so its units are response units per time unit. Two data values can determine unknown parameters, but a fitted curve is not proof of the mechanism.

Multiplicative comparisons become differences

On a model scale $S = k {log}_{10} (I / I_{0})$ , the input is a ratio of positive quantities. A difference in scale corresponds to an intensity ratio, not an intensity difference. State the model domain and the period or range where it is credible. Predictions far outside measured data need justification; a logarithm may keep increasing mathematically when a real response has a limit.

A common trap

A rise from 40 to 60 on a logarithmic scale means a 50% rise in the original quantity.

Instead: Use the supplied logarithmic formula. Scale differences describe ratios after exponentiation; they are not ordinary percentages of the intensity.

02 / WORKED EXAMPLES

See the thinking

Example 1

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.

$S (0) = a = 40$ .
$70 = 40 + b ln 4$ , so $b = 30 / ln 4$ .
$S (8) = 40 + 30 ln 9 / ln 4 \approx 87.55$ .

a = 40

b = 30 / ln 4

; day-8 score approximately

87.5

Example 2

Compare a logarithmic scale

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

Subtract the two equations: $15 = 10 {log}_{10} (I_{60} / I_{45})$ .
Thus $I_{60} / I_{45} = 10^{1.5}$ .
Calculate the ratio, rather than interpreting 15 as a percentage.

The 60-unit level has approximately

31.6

times the intensity.

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

$2$ score units per day.

$C^{'} (t) = 8 / (t + 1)$ , so $C^{'} (3) = 2$ .

03 / WORKSHEET

Your turn

6 questions · 17 marks · Work at your own pace.

Worksheet PDF 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 $S = 20 {log}_{10} (A / A_{0})$ . 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
$R^{'} (t) = 5 / (t + 1)$ , then substitute $t = 4$ .
$1$ response unit per hour.
A scale is $S = 20 {log}_{10} (A / A_{0})$ . What amplitude ratio gives a rise of 12 units?
3 marks
$12 = 20 {log}_{10} (A_{2} / A_{1})$ . 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.

Checking saved progress…

Ready for the topic test?

Review the chapter, then choose a practice test.

Topic review

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 ↗

Back to Logarithmic functions