AM-01.10 — Exponential and natural logarithm

Course compass

Question directrice : What do eˣ and ln do, how are they read, why are they inverse, and why does ln appear in the ideal rocket equation?

Free learning resource: this content is not professional training, a diploma, certification or professional authorization. Examples are educational and do not replace engineering, testing or qualification of real space hardware.

1 — Exponential change: multiply rather than add the same amount

If a reserve gains exactly 10 kg each hour, it grows by +10, +10, +10. Exponential growth is different: the change depends on what is already there. A culture that doubles goes 1, 2, 4, 8, 16.

Linear and exponential growth — Adding the same amount and multiplying by the same factor create different curves.

2 — What does eˣ mean and how is it read?

eˣ is read “e to the power x”. The number e is approximately 2.71828 and appears naturally in continuous-change models.

You do not need to memorize its digits. Think of eˣ as turning an exponent x into a multiplication factor.

3 — Natural logarithm: the inverse question

If we know the final factor and want the exponent that produced it, the natural logarithm, written ln, answers that question.

if eˣ = N, then x = ln(N)

Read: “if e to the power x equals N, then x equals the natural logarithm of N.”

Exponential and logarithm as inverses — The exponential goes from exponent to factor; the logarithm goes back from factor to exponent.

4 — Concrete example 1: how many doublings?

A culture goes 1 → 2 → 4 → 8. Reaching 8 takes three doublings because 2³=8.

Using logs: x = ln(8) ÷ ln(2) ≈ 2.079 ÷ 0.693 ≈ 3.

Why divide by ln(2)?

Because each step multiplies by 2. Dividing by ln(2) converts the natural-log measure into a number of doublings.

5 — Concrete example 2: why a logarithm appears in the ideal rocket equation

Δv = vₑ × ln(m₀ ÷ m₁)

Δv is read “delta vee” and is a change in velocity. vₑ is the effective exhaust velocity in this simplified model.

If m₀/m₁=2, ln(2)≈0.693. If the mass ratio becomes 4, ln(4)≈1.386. The benefit does not grow directly in proportion to mass ratio.

Logarithm in a rocket relation — The logarithm converts a multiplicative mass ratio into an additive velocity contribution.

6 — Why must the argument of ln be dimensionless?

In ln(m₀/m₁), both m₀ and m₁ are masses. If both use kilograms, kg/kg cancels, leaving a dimensionless ratio.

Writing ln(500 kg) without a reference quantity is not the same kind of meaningful dimensionless operation.

7 — Calculator: ln, eˣ and common mistakes

Compute ln(2): approximately 0.693. Then compute e^0.693 and you recover approximately 2.

Do not confuse ln and log₁₀. Many calculators provide both. They are different logarithms.

ln and exponential buttons — ln and eˣ undo each other.

8 — Three mental pictures

Exponential: from exponent to multiplication factor.
Logarithm: from factor back to exponent.
Rocket equation: a mass ratio contributes through a logarithm, so ever larger ratios become increasingly costly.

Exponential and logarithm summary — Start from the concrete question, then choose the function that answers it.

Exercises and answers

Exercise 1

A quantity doubles four times from 1. Final value?

2⁴=16.

Exercise 2

Compute ln(2), then e^ln(2).

ln(2)≈0.693; e^0.693≈2.

Exercise 3

Why is m₀/m₁ dimensionless if both masses are in kg?

Because kg/kg cancels.

AM-01.10 — Exponential and natural logarithm: repeated multiplication and its inverse