DWG NO. 3.2 — Lesson 2 of 6

The Number e and Exponential Growth/Decay

Unit 3: Exponential and Logarithmic Functions · ~20 min

In 3.1, the base b of an exponential function could be any positive number other than 1. One particular base shows up so often in growth and decay problems that it gets its own symbol: e ≈ 2.71828..., an irrational number sometimes called Euler's number. It isn't arbitrary — it's the base that falls out naturally when growth happens continuously rather than in discrete steps.

Where e comes from

Imagine money compounding more and more frequently — yearly, then monthly, then daily, then every second. As the number of compounding periods n grows without bound, the expression (1 + \(\frac{1}{n} )^{n}\) settles down and approaches a fixed value:

• n = 1: (1 + 1)¹ = 2
• n = 12: (1 + \(\frac{1}{12} )^{12}\) ≈ 2.613
• n = 365: (1 + \(\frac{1}{365} )^{365}\) ≈ 2.7146
• n → ∞: the expression approaches e ≈ 2.71828

That limiting value, e, is what appears whenever a quantity grows or decays continuously — smoothly, at every instant — rather than in fixed jumps.

The continuous growth/decay model

A quantity that grows or decays continuously follows:

• \(A_{0}\) — the initial amount, at t = 0.
• k > 0 — continuous growth (population increase, compounding money).
• k < 0 — continuous decay (radioactive decay, cooling, drug elimination).
• t — time, in whatever unit k was measured in.

This is the same shape as f(x) = a·\(b^{x}\) from 3.1, just written with base e and the rate built into the exponent as k instead of folded into b.

Guided practice

• 1.Between which two consecutive integers does e fall? Give e to two decimal places.
  Show answer

  e is between 2 and 3; to two decimal places, e ≈ 2.72.
• 2.In A(t) = \(A_{0} e^{kt}\), what does the sign of k tell you?
  Show answer

  k > 0 means continuous growth (the quantity increases over time); k < 0 means continuous decay (the quantity decreases over time).
• 3.An investment of $2000 grows continuously at k = 0.05 per year. Find its value after 8 years (\(e^{0.4}\) ≈ 1.4918).
  Show answer

  A(8) = 2000\(e^{0.05(8)}\) = 2000\(e^{0.4}\) ≈ 2000(1.4918) ≈ $2983.60.
• 4.A radioactive sample starts at 200 g and decays with k = −0.1 per year. Find the amount left after 4 years (\(e^{-0.4}\) ≈ 0.6703).
  Show answer

  A(4) = 200\(e^{-0.1(4)}\) = 200\(e^{-0.4}\) ≈ 200(0.6703) ≈ 134.1 g.
• 5.Why does \(e^{x}\) have the same domain, range, and asymptote as the other exponential graphs from 3.1, even though its base is irrational?
  Show answer

  e is just a specific positive number greater than 1 (e ≈ 2.718 > 1), so \(e^{x}\) is an ordinary growth curve of the form a·\(b^{x}\) with a = 1, b = e. Domain (−∞, ∞), range (0, ∞), asymptote y = 0 — every rule from 3.1 still applies.