# e, ln, odds and the sigmoid

> ln undoes a power, so it solves for time; odds = p ÷ (1 − p), log-odds and the sigmoid p = 1 ÷ (1 + e⁻ᶻ), eᵇ as an odds multiplier, and payback with churn.

- Course: [Math for Statistics and Unit Economics](https://productovi.com/courses/math-basics.md)
- Module: Growth and time
- Lesson 24/36, ~35 min
- Source: https://productovi.com/courses/math-basics/e-ln-odds-sigmoid

## ln undoes a power

**Alex Tutor:** Back at Klasna for the last lesson of the growth module. Two questions keep coming up: "how long until…?" and "how likely is…?". Both need two buttons from a calculator: e and ln.

**Sofia:** I know (1 + r)ᵗ. But how do I get t out of the exponent?

**Alex Tutor:** With a logarithm. The natural log, **ln**, turns a power into a product: ln(aᵗ) = t × ln a. So aᵗ = b gives t = ln b ÷ ln a.

$$
a^t = b \;\Rightarrow\; t = \frac{\ln b}{\ln a}
$$

*ln: the natural logarithm, a calculator or spreadsheet button (=LN). Any log works for this ratio, as long as both use the same one.*

**Alex Tutor:** How long does money take to double at 25% a year? 1.25ᵗ = 2, so t = ln 2 ÷ ln 1.25 = 0.6931 ÷ 0.2231 ≈ 3.1 years.

> **Exercise:** How many years does it take for money to triple at 8% a year? Two decimals.

## What e is

**Sofia:** And e? It shows up in every logistic formula.

**Alex Tutor:** e ≈ 2.718 is what 1 grows to at 100% a year if interest is added continuously, not once: (1 + 1/n)ⁿ as n grows. ln is simply the power of e: ln(eˣ) = x.

**Alex Tutor:** For this course you only need the pair: e to some power is a growth factor, and ln brings you back to the power.

**Sofia:** So eˣ and ln undo each other, like squaring and the square root.

**Alex Tutor:** A quick check on where e comes from.

> **Exercise:** e ≈ 2.718 is what 1 grows to in a year at 100% interest added once, at the end of the year.

## Odds and the sigmoid

**Alex Tutor:** Now "how likely". Say 20% of trial learners buy a package. The **odds** compare yes with no: 0.2 ÷ 0.8 = 0.25, "one to four". At p = 0.5 the odds are 1.

$$
\text{odds} = \frac{p}{1 - p} \qquad p = \frac{1}{1 + e^{-z}}, \quad z = \ln(\text{odds}) = b_0 + b_1 x
$$

*p: probability; z: the log-odds, the score a logistic regression adds up; the second formula, the sigmoid, turns any z into a p between 0 and 1.*

> **Exercise:** Three quarters of learners who attend a second lesson buy a package. What are the odds of buying?

**Alex Tutor:** And the way back, from odds to probability.

> **Exercise:** A model says the odds that a trial learner buys a package are 1.5. What is the probability of buying? Two decimals.

**Sofia:** Why take the log of the odds at all? Odds already compare yes with no.

**Alex Tutor:** Odds run from 0 to infinity and sit lopsided around 1: 0.25 and 4 are mirror images. ln makes them symmetric around 0, −1.39 and +1.39, and lets the model add things up.

**Alex Tutor:** A logistic model adds up a score z = ln(odds), for example z = −1.5 + 0.5 × lessons in the first week. Then the **sigmoid** turns z into a probability.

**Sofia:** With 3 lessons z = 0, so p = 1 ÷ (1 + e⁰) = 0.5. With 5 lessons z = 1: p = 1 ÷ (1 + e⁻¹) ≈ 0.73.

![The sigmoid turns a score z into a probability between 0 and 1: z = −1 gives 0.27, z = 0 gives 0.5, z = 1 gives 0.73; each step of 1 in z multiplies the odds by e ≈ 2.72.](https://productovi.com/courses/math-basics/diagrams/m6-l4-sigmoid.png)

*Steep in the middle, flat at the ends: the same step in z moves p most near 0.5.*

> **Exercise:** With z = −1.5 + 0.5 × lessons, what is the probability of buying for a learner with 1 lesson in the first week? Two decimals.

## eᵇ multiplies the odds

**Alex Tutor:** So what does b₁ = 0.5 mean? Not +0.5 in probability. Each extra lesson adds 0.5 to the log-odds, which multiplies the odds by e^0.5 ≈ 1.65.

**Sofia:** Because adding inside ln is multiplying outside. And the change in probability depends on where you start: big near 0.5, tiny near 0 or 1.

**Alex Tutor:** Try it: z = 0 gives odds of e⁰ = 1, z = 1 gives odds of e ≈ 2.72. One step in z, odds times e. With b₁ = 0.5, one lesson is half a step: times 1.65.

**Sofia:** So I should report "1.65 times the odds", not "+50% chance". Those are very different claims.

> **Exercise:** In a logistic model, the coefficient of "joined a speaking club" is b = 0.7. What does it mean?

## Payback with churn

**Alex Tutor:** Back to "how long". A Club member leaves 200 UAH a month, 5% leave each month, and getting one costs 1,500. After t months the expected margin is 200 × (1 − 0.95ᵗ) ÷ 0.05. When does it reach 1,500?

**Sofia:** Rearrange: 0.95ᵗ = 1 − 1,500 × 0.05 ÷ 200 = 0.625. Then ln: t = ln 0.625 ÷ ln 0.95 ≈ −0.470 ÷ −0.0513 ≈ 9.2 months.

$$
t = \frac{\ln\left(1 - \frac{CAC \cdot c}{g}\right)}{\ln(1 - c)}
$$

*g: margin per member per month; c: monthly churn; CAC: the cost to acquire one member. If CAC ≥ g ÷ c, the bracket is 0 or negative and the member never pays back.*

![Expected margin per member climbs towards 200 ÷ 0.05 = 4,000 UAH and crosses a 1,500 UAH acquisition cost after about 9.2 months; a cost above 4,000 would never be paid back.](https://productovi.com/courses/math-basics/diagrams/m6-l4-payback-with-churn.png)

*The curve flattens because members leave, so payback can be "never".*

**Alex Tutor:** And the cap is 4,000, the LTV from last lesson. Pay more than that and the formula asks for the log of a negative number: there is no such month.

**Alex Tutor:** Try the payback formula on other numbers.

> **Exercise:** A Club member leaves 250 UAH of margin a month, monthly churn is 5% and acquiring one member costs 2,000 UAH. After how many months does the expected margin pay back the cost? Two decimals.

> **Exercise:** A colleague reads the speaking-club model (b = 0.7) and writes: "Joining a club raises the chance of buying by 70 percentage points." Correct this in 2–3 sentences, with one example.

## Case: which channel pays back?

**Alex Tutor:** Last task of the module. Marketing compares two channels for the Club, and Leo's rule is simple: a member must pay back within 12 months.

**Sofia:** So payback for each with the ln formula, and the cap g ÷ c to see if it's possible at all.

**Alex Tutor:** And remember the two channels bring different members: churn differs, not just the price of getting them.

> **Exercise:** Compute payback for both channels and recommend which one passes Leo's 12-month rule.

## Where you will meet this

> **In short**
>
> - **ln** undoes a power: aᵗ = b ⇒ t = ln b ÷ ln a. Doubling at 25% takes ln 2 ÷ ln 1.25 ≈ 3.1 years.
> - **e** ≈ 2.718; eˣ and ln undo each other.
> - **Odds** = p ÷ (1 − p); a logistic model adds up z = ln(odds), and the **sigmoid** p = 1 ÷ (1 + e⁻ᶻ) turns it back into a probability.
> - A coefficient b multiplies the odds by **eᵇ**; the change in probability depends on the starting point.
> - Payback with churn: t = ln(1 − CAC·c ÷ g) ÷ ln(1 − c); if CAC ≥ g ÷ c, never.

> **These lessons use it**
>
> - [Logistic regression and classification](/courses/statistics-for-pm-and-marketers/logistic-regression): odds, ln(odds) = b₀ + b₁x and the sigmoid.
> - [CAC, payback and ROI](/courses/product-analytics/cac-payback-roi): when a cohort pays back its acquisition cost.
> - [Discounted LTV and CAC payback in cash](/courses/finance-for-product/ltv-and-cac-payback-in-cash): payback month by month.

**Alex Tutor:** That closes growth and time. Next, probability itself: p, 1 − p and what they hide.

- Previous lesson: [Geometric series](https://productovi.com/courses/math-basics/geometric-series.md)
- Next lesson: [p and 1 − p](https://productovi.com/courses/math-basics/p-and-1-minus-p.md)

---
Productovi Learning. Free, no sign-up needed to read.