IB Maths AI HL Log, Logistic & Piecewise Paper 2 & 3 a + b ln x, monotonic ~9 min read

Natural Logarithmic Models

A natural logarithmic model has the form f(x) = a + b ln x for x > 0. It models things that increase rapidly at first then keep growing without ever stopping — sound decibels, earthquake magnitudes, tree heights. The graph always passes through (1, a) (since ln 1 = 0) and has a vertical asymptote at the y-axis.

📘 What you need to know

Standard form: f(x) = a + b ln x, x > 0. (ln x ≡ log_e x.)
Always passes through (1, a): a = value of the function when x = 1 — the easiest data point to read off.
b controls direction and rate: b > 0 increasing, b < 0 decreasing; larger |b| means faster change.
Vertical asymptote: x = 0 (the y-axis). No y-intercept.
One root at x = e^−a/b, found by solving a + b ln x = 0.
Monotonic, no max or min: the curve only goes one direction (up or down).
Unbounded: as x → ∞ the function keeps growing (or shrinking) without limit — that’s a key real-world limitation.

The shape of a + b ln x

The natural logarithm grows fast near 1 but slows down dramatically — doubling from x = 1 to x = 2 raises ln by 0.69, but doubling from x = 100 to x = 200 raises ln by the same 0.69. That’s why log models are perfect for quantities that “feel” multiplicative: every 10× increase in sound intensity adds the same number of decibels; every 10× increase in earthquake energy raises the magnitude by one unit. The vertical asymptote at x = 0 means the model breaks down as the input approaches zero — ln 0 is undefined and the function dives to −∞.

Both curves share a = 4 (passing through (1, 4)) but differ in sign of b. The teal curve (b = 2) increases; the blue curve (b = −2) decreases. Both have the y-axis as a vertical asymptote.

Natural log model at a glance f(x) = a + b ln x, x > 0 f(1) = a · root at x = e^−a/b · vertical asymptote x = 0 · monotonic

Finding a and b from data

Two data points are enough to pin down both parameters. The smart play: if one of the data points is at x = 1, then ln 1 = 0 instantly gives you a; substitute the second point to solve for b. If neither point is at x = 1, you get a 2×2 linear system in a and b — subtract one equation from the other to eliminate a. Once you have the model, the GDC lets you solve for any unknown value, forward or backward.

x = 1 is your friend: any time a logarithmic model has a data point at x = 1, the constant a falls out instantly because ln 1 = 0. Look for it first.

When does a log model fit?

Log models suit phenomena that grow (or shrink) without limit but at a slowing rate: each additional unit of input produces a smaller and smaller effect on output. Classic applications include perceived loudness (decibels), perceived brightness (stellar magnitudes), earthquake magnitudes, biological growth where resources don’t limit individual size, and learning/memory curves. The main limitation: real-world quantities usually have a ceiling or floor that log models can’t represent — for those, you may need a logistic model instead.

🧭 Recipe — natural log models

Identify a: if you know f(1), then a = f(1) directly.
Identify b: substitute a second data point and solve.
Forward calculation: to find f(x) at a given x, substitute into a + b ln x.
Backward calculation: to find x for a given f(x), rearrange to ln x = (f − a)/b and apply e^… (or use GDC’s solver).
Check the context: state the domain (usually x > 0) and mention if the prediction is reliable given any real-world limits.

Worked examples

WE 1

Key features of a log graph

Given f(x) = 4 + 2 ln x: (a) State the point through which the graph passes when x = 1. (b) State the equation of the vertical asymptote. (c) Is the function increasing or decreasing? (d) Find the root.

(a) substitute x = 1 f(1) = 4 + 2 ln 1 = 4 + 0 = 4 passes through (1, 4) (b) asymptote at y-axis x = 0 (c) sign of b b = 2 > 0 → increasing (d) solve f(x) = 0 4 + 2 ln x = 0 ⇒ ln x = −2 x = e⁻² ≈ 0.135

WE 2

Find a and b from two data points

A natural log model N(x) = a + b ln x satisfies N(1) = 30 and N(e²) = 42. (a) Find a and b. (b) Find N(10).

(a) use N(1) first N(1) = a + b ln 1 = a = 30 a = 30 substitute into N(e²) 30 + b ln e² = 42 30 + 2b = 42 ⇒ b = 6 N(x) = 30 + 6 ln x (b) substitute x = 10 N(10) = 30 + 6 ln 10 ≈ 30 + 6(2.3026) N(10) ≈ 43.82 (4 sf)

WE 3

Decreasing model: reaction time

The reaction completion time (minutes) of a chemical process is modelled by T(c) = 50 − 6 ln c, where c is the catalyst concentration (mg/L). (a) Find T(1). (b) Find T(20). (c) Find the concentration when the reaction time is 20 min.

(a) T(1) = 50 − 6 ln 1 T(1) = 50 min (b) T(20) = 50 − 6 ln 20 = 50 − 6(2.9957) T(20) ≈ 32.03 min (c) solve 50 − 6 ln c = 20 6 ln c = 30 ⇒ ln c = 5 c = e⁵ ≈ 148.4 mg/L b = −6 < 0, so the time decreases as concentration increases (faster reaction). ✓

WE 4

Tree growth: find parameters and predict age

A tree’s height H(t) metres at age t years is modelled by H(t) = a + b ln t. Measurements give H(1) = 2 m and H(8) = 5 m. (a) Find a and b. (b) Find the age when the tree first reaches 7 m.

(a) H(1) = a = 2 a = 2 substitute into H(8) 2 + b ln 8 = 5 ⇒ b ln 8 = 3 b = 3/ln 8 ≈ 1.443 (b) solve H(t) = 7 2 + (3/ln 8) ln t = 7 ln t = (5 · ln 8)/3 = (5/3) ln 8 t = e^[(5/3) ln 8] = 8^(5/3) = (8^(1/3))⁵ = 2⁵ t = 32 years (exact) clean exact answer thanks to 8 = 2³.

WE 5

Earthquake magnitude

The magnitude M of an earthquake with relative intensity I is modelled by M(I) = 4 + 0.5 ln I. (a) Find M(1). (b) Find M(100) (2 d.p.). (c) Find the intensity of an earthquake of magnitude 7 (exact form, then 1 d.p.).

(a) M(1) = 4 + 0.5 ln 1 M(1) = 4 (b) M(100) = 4 + 0.5 ln 100 = 4 + 0.5(4.6052) M(100) ≈ 6.30 (c) solve M(I) = 7 4 + 0.5 ln I = 7 0.5 ln I = 3 ⇒ ln I = 6 I = e⁶ ≈ 403.4

WE 6

Memory retention — limitations

The percentage of new information a student retains after t days without review is modelled by R(t) = 80 − 10 ln t, valid for t ≥ 1. (a) Find R(1) and R(7). (b) After how many days does retention drop to 50%? (c) State one limitation of this model.

(a) R(1) = 80 − 10 ln 1 R(1) = 80% R(7) = 80 − 10 ln 7 ≈ 80 − 19.46 R(7) ≈ 60.5% (b) solve R(t) = 50 80 − 10 ln t = 50 ⇒ ln t = 3 t = e³ ≈ 20.1 days (c) limitation R is unbounded below: for t > e⁸ ≈ 2981 days, the model gives R < 0, which is impossible (retention can’t be negative). The model fails for very large t.

💡 Top tips

Substitute x = 1 first: ln 1 = 0 collapses the model to f(1) = a, giving you a in one step.
Sign of b tells you the direction: b > 0 means rising; b < 0 means falling.
The graph never touches x = 0 — that’s the vertical asymptote. Domain is x > 0.
Use your GDC’s solver for backward calculations (finding x from f) when the algebra gets messy.
Watch for clean exact answers: powers of e, or expressions that simplify like 8^5/3 = 32.

⚠ Common mistakes

Treating ln as log₁₀ — in this course ln means log_e. Use the GDC’s “ln” key, not “log”.
Forgetting ln 1 = 0 — missing the obvious step that immediately gives a.
Extrapolating beyond the model’s domain — predictions for very small or very large x are often nonsense.
Confusing increasing/decreasing — the sign of b alone decides it.
Forgetting the asymptote — there’s no y-intercept; the curve runs off to −∞ (or +∞) as x → 0⁺.

Next up: Logistic Models — when growth eventually levels off at a carrying capacity, log alone won’t do. The S-shaped logistic curve handles bounded growth.

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