TwiddleLab / Math / Growth: Exponents and Logs
Faster and Faster: the Number e
What you will learn
- Compound interest pays interest on interest: 50% twice turns $1 into $2.25, not $2.
- Splitting 100% into n payments gives (1 + 1/n)ⁿ, which approaches e ≈ 2.71828.
- Growth paid continuously is e^(rt). Its inverse is the natural log ln: ln 2 ≈ 0.69, so 6.9% doubles in 10 years.
How the lesson goes
A bank pays 100% a year. Would you rather get it once, or 1% a hundred times?
Once turns $1 into $2. Bet what 1% a hundred times gives.
Start small: two payments of 50%.
Pay the interest twice and watch the stack.
Split the 100% into more and more payments.
Turn the dial from once a year to every second.
The ceiling is a number: e.
(1 + 1/n)ⁿ grows with n but never passes e ≈ 2.71828.
Growth paid continuously at rate r for time t multiplies by e^(rt).
The inverse of eˣ is the natural log, ln x: ln e = 1, ln 2 ≈ 0.693.
Make $1 double in exactly 10 years.
The bank pays continuously, so $1 becomes e^(rt). Turn the rate until it lands on $2.
Doubling means e^(10r) = 2. The natural log undoes e: 10r = ln 2 ≈ 0.693, so r ≈ 6.9%. That is where the rule of 70 comes from: 70 / rate ≈ years to double.
Game: beat the bank.
Four rounds. Tap the bank that turns $1,000 into more in one year.
AI uses e in the sigmoid.
The sigmoid 1 / (1 + e^(−x)) squeezes any number into 0 to 1. Drag the point until the output is 0.9.
1/(1 + e^(−x)) = 0.9 when e^(−x) = 1/9, so x = ln 9 ≈ 2.2.
$1,000 at 100% a year, paid continuously. How much after one year?
Paying 100% continuously multiplies by e^(1 × 1) = e ≈ 2.71828, so $1,000 becomes about $2,718.28.
One more thing that makes e special.
Tap a curve to see its tangent at x = 0. Which one has slope exactly 1? Then Check.
The slope of bˣ at 0 is ln b: ln 2 ≈ 0.69, ln 3 ≈ 1.10, and ln e = 1. That is why e is the natural base.
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