A principal P is invested at a nominal annual rate r, compounded n times a year, for t years. Every quantity in that sentence has a unit attached, and most errors with these formulas are unit errors rather than algebra errors.
Definition · Nominal rateThe nominal annual rate r is quoted per year and split evenly among the n periods, so one period carries the rate r/n. A nominal 5% compounded monthly means 5/12 of a percent a month. It is a name for twelve times the monthly rate, not the growth actually achieved over a year.
The first formula is a repeated multiplication and nothing more. Each period multiplies the balance by 1 + r/n, so the balances form a geometric sequence.
after 1 period P(1 + r/n)
after 2 periods P(1 + r/n)²
after k periods P(1 + r/n)^k
t years is nt periods, so A = P(1 + r/n)^(nt)
Definition · Effective annual rateThe effective annual rate is (1 + r/n)^n − 1: the single yearly rate that would produce the same balance after one year. For a nominal 5% compounded monthly it is (1 + 0.05/12)^12 − 1 ≈ 0.051162, that is 5.1162%. Two accounts quoted at different frequencies can only be compared through this number.
Now increase n and watch. The numbers below are $1000 at a nominal 5% for 10 years, each rounded to the nearest cent.
n = 1 yearly 1000(1.05)^10 = 1628.89
n = 2 half-yearly 1000(1.025)^20 = 1638.62
n = 4 quarterly 1000(1.0125)^40 = 1643.62
n = 12 monthly 1000(1 + 0.05/12)^120 = 1647.01
n = 365 daily 1000(1 + 0.05/365)^3650 = 1648.66
n → ∞ continuous 1000 · e^0.5 = 1648.72
simple interest, for scale 1000(1 + 0.05 · 10) = 1500.00
The column climbs and then stops climbing. Daily and continuous differ by about six cents over ten years on a thousand dollars. "Compounded continuously" is not a claim about speed — it is the ceiling of the process, and the last few frequencies are already sitting against it.
Theorem · The continuous limitAs n → ∞, P(1 + r/n)^(nt) → P · e^(rt).
Substitute m = n/r, so that r/n = 1/m and nt = m · rt. The expression becomes P[(1 + 1/m)^m]^(rt), and m → ∞ as n → ∞. The inner bracket tends to e by “The Number e”, so the whole expression tends to P · e^(rt).
Solving for the other letters
P and A come out by dividing. r and t sit in the exponent, so they need “Solving Exponential Equations”. To double a balance under continuous growth, set 2P = P · e^(rt); the P cancels, 2 = e^(rt), and rt = ln 2, so t = ln 2 / r.
That the principal cancels is the characteristic property of exponential growth from “Growth and Decay”: the time to double does not depend on where you start. At 5% continuous it is ln 2 / 0.05 ≈ 13.86 years, and compounded yearly instead it is ln 2 / ln(1.05) ≈ 14.21 years.
The banker's shortcut follows from ln 2 ≈ 0.6931. Doubling time is roughly 70 divided by the rate as a percent, which at 5% gives 14 years against a true 13.86. It is an approximation of a logarithm, and it drifts at large rates: at 20% the rule says 3.5 years and the continuous answer is 3.47, but at 100% it says 0.7 and the answer is 0.69.
Example · What a percent sign costsEntering 5 instead of 0.05 for a monthly-compounded account makes the monthly factor 1 + 5/12, a rate of 500% a year. Over ten years $1000 becomes about 1.4 · 10^21 dollars. The formula reports it without complaint, because nothing in the algebra knows what a percent is.