Exponential regression | Year 12

01 / LEARN

The idea, step by step

Recognise multiplicative change

An exponential model $y = A b^{x}$ has a positive starting scale A and positive growth factor b. Equal increments in x multiply the predicted response by the same factor. Values b>1 give growth; 0<b<1 give decay. Positive response data can be transformed using logarithms to fit an exponential model.

A straight line fitted to ln(y) against x gives $ln (\hat{y}) = α + β x$ , so $\hat{y} = e^{α} {(e^{β})}^{x}$ . The intercept of the log line is not A itself.

ln (\hat{y}) = α + β x \Rightarrow \hat{y} = e^{α} e^{β x}

Check predictions on the original scale

Plot the original data and inspect prediction errors in original response units. An exponential fit to logged values minimises squared log residuals, so it is not the same fitting criterion as least squares in the untransformed y-values.

A logarithm-based fit requires positive responses. Treat a logged r² as describing variation on the log scale; do not automatically call it the percentage of original y-variation explained. State the observed x-range and avoid unjustified long-term forecasts.

A common trap

The intercept 2.3 in ln(ŷ)=2.3-0.08x means y starts at 2.3.

Instead: Exponentiate the entire equation: the original-scale starting value is $e^{2.3}$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

An exact growth model

A culture has counts 80,100,125 at x=0,1,2 hours. Find an exponential model.

Paired measurements
Time (h)	Count
0	80
1	100
2	125

The ratios are 100/80=1.25 and 125/100=1.25.
The initial value is 80 and the hourly multiplier is 1.25.
At hour 3 the model predicts 80×1.25³=156.25.

\hat{y} = 80 {(1.25)}^{x}

; predicted count 156.25 before sensible rounding.

Example 2

Undo a log fit

Technology gives ln(ŷ)=2.3-0.08x. Write the exponential model and interpret its multiplier.

The initial scale is $e^{2.3}$ ≈9.9742.
The multiplier is $e^{- 0.08}$ ≈0.9231.
The response falls by about 7.69% per x-unit.

\hat{y} \approx 9.9742 {(0.9231)}^{x}

Check your understanding

A fitted model is $\hat{y} = 200 {(0.9)}^{x}$ . What percentage change occurs each x-unit?

Show answer

A 10% decrease.

The multiplier retains 90% of the previous modelled amount.

03 / WORKSHEET

Your turn

6 questions · 18 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Find an exact exponential model for y=50,100,200 at x=0,1,2.
3 marks
For $\hat{y} = 120 {(1.1)}^{x}$ , predict the response at x=3.
3 marks
A log fit is ln(ŷ)=1.5+0.2x. Find A and b in $\hat{y} = A b^{x}$ .
3 marks
Why cannot a zero or negative response be entered directly into a real natural-log regression?
3 marks
A model multiplies by 0.96 each week. Find its modelled change over three weeks.
3 marks
A log-scale regression has r²=0.94. State a careful interpretation.
3 marks

Read the worked solutions

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

Find an exact exponential model for y=50,100,200 at x=0,1,2.
3 marks
Successive values have ratio 2.
At x=0 the response is 50.
$y = 50 \times 2^{x}$ .
For $\hat{y} = 120 {(1.1)}^{x}$ , predict the response at x=3.
3 marks
Calculate 120×1.1³.
159.72.
A log fit is ln(ŷ)=1.5+0.2x. Find A and b in $\hat{y} = A b^{x}$ .
3 marks
Exponentiate the intercept and slope separately.
$A = e^{1.5}$ and $b = e^{0.2}$ .
A≈4.4817, b≈1.2214.
Why cannot a zero or negative response be entered directly into a real natural-log regression?
3 marks
A different justified model or transformation is needed.
Do not silently discard values.
The real logarithm ln(y) is defined only for y>0.
A model multiplies by 0.96 each week. Find its modelled change over three weeks.
3 marks
Calculate 0.96³.
Percentage decrease=100(1-0.884736).
Multiplier 0.884736, so a decrease of about 11.53%.
A log-scale regression has r²=0.94. State a careful interpretation.
3 marks
The statistic belongs to the transformed response.
It is not automatically 94% of original y-variation or evidence of causation.
94% of the observed variation in ln(y) is accounted for by its fitted linear relationship with x.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 General Mathematics: statistical models. Original lessons and assessment practice; not an official SACE assessment.

Official curriculum source ↗

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