e and Natural Logarithms
The number that shows up whenever growth is continuous.
The explanation
e is a specific number, about 2.71828, like π is about 3.14159.
It appears whenever growth is continuous rather than in steps. If interest compounds every instant instead of monthly, the formula becomes A = Pe^(rt).
ln means log base e. It is the inverse of eˣ, so:
ln(eˣ) = x and e^(ln x) = x.
Where does e come from? Compound $1 at 100% for a year. Yearly gives $2. Monthly gives $2.61. Daily gives $2.7146. Every instant gives e.
All the log properties apply to ln unchanged. It is a logarithm like any other, with a base that happens to be irrational.
The constant is defined by e = lim_{n→∞}(1 + 1/n)^n ≈ 2.718281828, equivalently by the series Σ 1/k!.
Continuous compounding follows from that limit: A = P(1 + r/n)^{nt} → Pe^{rt} as n → ∞. The same form y = y₀e^{kt} models any process whose rate of change is proportional to its current amount, with k > 0 for growth and k < 0 for decay.
The defining property is that e^x is its own derivative, which is why it becomes the standard base once calculus begins; d/dx(ln x) = 1/x likewise makes ln the natural antiderivative choice.
Any exponential can be rewritten with base e via b^x = e^{x ln b}, so the general model a·b^t and the continuous model a·e^{kt} are the same family with k = ln b. Converting between a stated percentage rate and a continuous rate uses exactly this identity, and the two are close for small rates but diverge as the rate grows.
Worked example
$2,000 is invested at 4.5% compounded continuously. Find the value after 7 years, and the time to double.
- A = 2000e^(0.045×7) = 2000e^0.315.
- e^0.315 ≈ 1.3702, so A ≈ $2,740.42.
- Doubling: 2 = e^(0.045t), so ln 2 = 0.045t.
- t = 0.6931/0.045.
Answer: About $2,740.42; doubling takes about 15.4 years.
Common mistakes
- Treating e as a variable rather than a constant.
- Using the annual formula A = P(1 + r)^t when the problem says continuously.