MISSION · Algebra
e and the Natural Logarithm
The one base where the growth rate equals the amount you already have
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The idea in one sentence
\(e \approx 2.71828\) is the base for which the graph of \(y = e^x\) has a gradient equal to its own height at every point — so the thing grows at exactly the rate of what is already there.
Where the number comes from
Put $1 in an account paying 100% interest a year. After a year you have $2.
Now pay it half-yearly at 50% each time: \((1 + \tfrac12)^2 = 2.25\). Monthly at \(\tfrac{1}{12}\) each: \((1 + \tfrac{1}{12})^{12} \approx 2.613\). Daily: \(\approx 2.7146\). Every second: \(\approx 2.71828\).
More frequent compounding pays more, because interest starts earning interest sooner — but the gain shrinks each time, and the total closes in on a ceiling rather than running away.
\[e = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n \approx 2.718281828\]That ceiling is \(e\). It is what compound interest becomes when the compounding never stops, and it is irrational — the repeating 1828 is a coincidence, and the digits never settle.
Why this base and not another
Every exponential curve \(y = a^x\) has a gradient proportional to its height. What makes \(e\) special is that the constant of proportionality is exactly 1:
\[\frac{d}{dx}e^x = e^x\]For \(2^x\) the gradient is about \(0.693\) times the height; for \(3^x\), about \(1.099\) times. Somewhere between 2 and 3 sits the base where the factor is 1, and that base is \(e\).
This is the only reason \(e\) appears everywhere in calculus. Any exponential can be rewritten as \(a^x = e^{x\ln a}\), so the whole family reduces to one function that differentiates to itself.
The natural logarithm
\(\ln x\) means \(\log_e x\) — the power you must raise \(e\) to in order to get \(x\). It undoes \(e^x\), and \(e^x\) undoes it:
\[\ln(e^5) = 5 \qquad e^{\ln 7} = 7\]Every logarithm law holds for \(\ln\) unchanged, because it is an ordinary log that happens to have an awkward base. And its derivative is as clean as you could hope:
\[\frac{d}{dx}\ln x = \frac{1}{x}\]The continuous growth model
Anything growing at a rate proportional to its own size follows
\[A = A_0 e^{kt}\]where \(A_0\) is the starting amount, \(k\) the continuous rate, and \(t\) the time. Positive \(k\) grows, negative \(k\) decays.
To solve for the time, take \(\ln\) of both sides. Bacteria doubling continuously at \(k = 0.35\) per hour, starting at 500 — when do they reach 5000?
\[10 = e^{0.35t} \; \Rightarrow \; \ln 10 = 0.35t \; \Rightarrow \; t = \frac{2.303}{0.35} \approx 6.6 \text{ hours}\]The same model handles radioactive decay, drug concentration in blood, cooling coffee, charging capacitors, and continuously compounded returns. Once the rate of change is proportional to the amount, \(e\) is not a choice — it is what the differential equation produces.
Common mix-ups
- Treating \(e\) as a variable. It is a fixed number, like \(\pi\).
- Confusing \(\ln\) and \(\log\). \(\ln\) is base \(e\); \(\log\) on a calculator is base 10. They differ by a constant factor, so swapping them scales every answer by 2.303.
- Writing \(\ln(a + b) = \ln a + \ln b\). The laws cover products.
- Reading \(e^{-x}\) as negative. It is always positive — small, but positive.
- Mixing up \(k\) with an annual percentage. A continuous rate of 0.05 is slightly more than 5% a year, because \(e^{0.05} \approx 1.0513\).
Check yourself
Give \(e\) to three decimal places, evaluate \(\ln(e^5)\), and solve \(e^x = 20\).
\(e \approx 2.718\). Since \(\ln\) undoes \(e^x\), \(\ln(e^5) = 5\). And \(x = \ln 20 \approx 3.00\).
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