📌 What Is a Semi-log Plot?
A semi-log plot has one axis scaled logarithmically. In AP Precalculus, we only scale the vertical (y) axis. The x-axis stays linear (equally spaced).
🔑 Key Fact: On a semi-log plot with a logarithmically scaled y-axis,
exponential functions appear linear.
exponential functions appear linear.
Equally-spaced y-values on the log scale are proportional (e.g. 1, 10, 100, 1000). The x-axis stays equally spaced.
📊 Normal Scale vs Semi-log Scale
Normal Scale
Exponential → curved
Semi-log Scale
Exponential → straight line ✓
✏️ Example 1 — Justifying an Exponential Model
Population of English Americans (1620–1820)
The left graph (normal scale) shows a curved pattern — indistinguishable for early years. The right graph (semi-log) appears linear after about t = 40 (year 1660).
Justification: If the semi-log plot appears linear, the data is exponential. We can use this observation to justify choosing an exponential model.
✏️ Example 2 — Identifying the Exponential Function from a Semi-log Plot
Math joke spreading — semi-log plot is linear
The semi-log plot is linear, so the underlying model is exponential: P(t) = ab^t.
The line on a semi-log plot has the form:
log P = b + mt (linear in log P)
→ P(t) = 10^(b+mt) = 10^b · (10^m)^t
This matches P(t) = a · b^t where:
a = 10^b and b_base = 10^m
Answer: D — P(t) = 2(2)^t
(if b = m and 10^b = 2)
log P = b + mt (linear in log P)
→ P(t) = 10^(b+mt) = 10^b · (10^m)^t
This matches P(t) = a · b^t where:
a = 10^b and b_base = 10^m
Answer: D — P(t) = 2(2)^t
(if b = m and 10^b = 2)
⚠️ Important Note — What "Logarithmically Scaled" Means
We do NOT change the actual y-values!
Logarithmically scaling the y-axis means equally-spaced gridlines correspond to proportional (multiplicative) y-values — like 1, 10, 100, 1000. The data points keep their original y-values.
Normal y-axis spacing: 0, 1, 2, 3, 4 → additive (+1 each step)
Log y-axis spacing: 1, 10, 100, 1000 → multiplicative (×10 each step)
A point with y = 50 still has y = 50 — it just plots between the 10 and 100 gridlines.
Log y-axis spacing: 1, 10, 100, 1000 → multiplicative (×10 each step)
A point with y = 50 still has y = 50 — it just plots between the 10 and 100 gridlines.
✏️ Example 3 — Identifying the Correct Semi-log Graph
f: x = 1,2,3,4,5 and f(x) = 40,60,90,135,203
| x | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| f(x) | 40 | 60 | 90 | 135 | 203 |
| log f(x) | 1.602 | 1.778 | 1.954 | 2.130 | 2.307 |
The log values increase by ≈ 0.176 each step → linear on a semi-log plot.
The correct graph must have:
• x-axis: equally spaced (linear)
• y-axis: log scale (gridlines at 1, 10, 100, 1000 — powers of 10)
• Points at y = 40, 60, 90, 135, 203 plotted between log gridlines
Answer: Graph D — y-axis labeled 1, 10, 100, 1000 with points correctly placed between gridlines.
The correct graph must have:
• x-axis: equally spaced (linear)
• y-axis: log scale (gridlines at 1, 10, 100, 1000 — powers of 10)
• Points at y = 40, 60, 90, 135, 203 plotted between log gridlines
Answer: Graph D — y-axis labeled 1, 10, 100, 1000 with points correctly placed between gridlines.
📐 Linear Models for a Semi-log Plot
Exponential model: y = ab^x
↕
Linear semi-log model: y = (log_n b)x + log_n a
Slope = log_n b | y-intercept = log_n a | n = base of the log scale
✏️ Example 5 — Find Linear Model, Then Exponential Model
Using f(x) = 40,60,90,135,203 data
a) Find the linear model y = mx + c (for the semi-log plot):
AROC (slope) = (log 60 − log 40) / (2 − 1) = 0.17609…
Find y-intercept: use point (1, log 40):
log a = log 40 − 0.17609(1) = 1.602 − 0.17609 ≈ 1.4259
Linear model: y = 0.17609x + 1.4259
─────────────────────────
b) Convert to exponential model y = ab^x:
a = 10^(y-intercept) = 10^1.4259 ≈ 26.67
b = 10^(slope) = 10^0.17609 ≈ 1.5
Exponential model: y = (80/3)(1.5)^x ≈ 26.67(1.5)^x
AROC (slope) = (log 60 − log 40) / (2 − 1) = 0.17609…
Find y-intercept: use point (1, log 40):
log a = log 40 − 0.17609(1) = 1.602 − 0.17609 ≈ 1.4259
Linear model: y = 0.17609x + 1.4259
─────────────────────────
b) Convert to exponential model y = ab^x:
a = 10^(y-intercept) = 10^1.4259 ≈ 26.67
b = 10^(slope) = 10^0.17609 ≈ 1.5
Exponential model: y = (80/3)(1.5)^x ≈ 26.67(1.5)^x
Connection: slope of semi-log linear model = log_n(b), y-intercept = log_n(a). To recover a and b: raise n to each value.
Card 1 of 7
Concept
What is a semi-log plot and which axis is scaled logarithmically in AP Precalculus?
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Answer
A plot where one axis is log-scaled — in AP Precalculus, always the vertical (y) axis
The x-axis stays equally spaced (linear). The y-axis has gridlines at proportional values: 1, 10, 100, 1000… The actual y-values of data points do NOT change.
Topic 2.15
Semi-log Plots
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Question 1 of 5
What is the key property of a semi-log plot with a logarithmically scaled y-axis?
Question 2 of 5
On a semi-log plot, the semi-log plot appears linear. Which function type is appropriate for the data?
Question 3 of 5
What does "logarithmically scaling the y-axis" mean for a semi-log plot?
Question 4 of 5
A semi-log plot has a linear model with slope m = 0.17609 and y-intercept = 1.4259 (base 10 log scale). What is the base b of the exponential model y = ab^x?
slope = log₁₀(b) → b = 10^(slope)
Question 5 of 5
A semi-log linear model has slope = log₁₀(b) and y-intercept = log₁₀(a). If y-intercept = 1.4259, what is a?
a = 10^(y-intercept) = 10^1.4259
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