High School Mathematics · Grades 10–12

23Exponential and Logarithm

The exponential function is the unique function equal to its own derivative and taking the value $1$ at $0$ . It converts sums into products; its inverse, the natural logarithm, converts products into sums. Together they describe every phenomenon whose rate of change is proportional to its size: radioactive decay, population growth, compound interest.

23.1 The exponential function

Theorem 23.1 (Existence and uniqueness)

There exists a unique differentiable function $f \colon \R \to \R$ such that

f' = f \qquad\text{and}\qquad f(0) = 1 .

It is called the exponential function and written $\exp$ , or $x \mapsto \eu^x$ .

Proof of uniqueness. First, such a function never vanishes. Indeed, let $g(x) = f(x)f(-x)$ ; then

g'(x) = f'(x)f(-x) - f(x)f'(-x) = f(x)f(-x) - f(x)f(-x) = 0,

so $g$ is constant equal to $g(0) = 1$ : for every $x$ , $f(x)f(-x) = 1$ , and in particular $f(x) \neq 0$ .

Now let $f_1, f_2$ be two solutions and $h = \dfrac{f_1}{f_2}$ (legitimate since $f_2$ never vanishes). Then

h' = \frac{f_1' f_2 - f_1 f_2'}{f_2^2} = \frac{f_1 f_2 - f_1 f_2}{f_2^2} = 0,

so $h$ is constant equal to $h(0) = 1$ , i.e. $f_1 = f_2$ .

Existence is admitted at this level (it can be obtained via the sequence of Exercise 20.10, or as the inverse of the logarithm constructed by integration in Chapter 25). ∎

Proposition 23.2 (Functional equation)

For all $x, y \in \R$ and $n \in \Z$ :

\eu^{x+y} = \eu^x\,\eu^y, \qquad \eu^{-x} = \frac{1}{\eu^x}, \qquad \eu^{x-y} = \frac{\eu^x}{\eu^y}, \qquad \bigl(\eu^{x}\bigr)^n = \eu^{nx}.

Moreover $\eu^x > 0$ for all $x$ .

Proof. Fix $y$ and consider $\varphi(x) = \dfrac{\exp(x+y)}{\exp(x)\exp(y)}$ . Its numerator and denominator, as functions of $x$ , are both solutions of $f' = f$ up to the constant $\exp(y)$ ; differentiating $\varphi$ directly (quotient rule) gives $\varphi' = 0$ , so $\varphi \equiv \varphi(0) = 1$ , proving the first identity. Taking $y = -x$ gives the second (with the value $\eu^0 = 1$ ), and the third follows. The fourth is an induction from the first for $n \geq 0$ , extended to $n < 0$ by the second.

Positivity: $\eu^x = \left(\eu^{x/2}\right)^2 \geq 0$ and $\eu^x \neq 0$ (shown in Theorem 23.1), so $\eu^x > 0$ . ∎

Proposition 23.3 (Variations and limits)

The exponential is strictly increasing, convex, and

\lim_{x\to-\infty} \eu^x = 0, \qquad \lim_{x\to+\infty} \eu^x = +\infty .

Proof. $(\exp)' = \exp > 0$ gives strict increase; $(\exp)'' = \exp > 0$ gives convexity. By convexity, $\eu^x \geq 1 + x$ (tangent at $0$ , see Example 22.15), so $\eu^x \to +\infty$ as $x \to +\infty$ by comparison. Then $\eu^{x} = 1/\eu^{-x} \to 0$ as $x \to -\infty$ . ∎

The exponential: strictly increasing, convex, with _-∈fty x = 0. It lies above its tangent at 0: x ≥ 1 + x. — The exponential: strictly increasing, convex, with $\lim_{-\infty} \eu^x = 0$ . It lies above its tangent at $0$ : $\eu^x \geq 1 + x$ .

Theorem 23.4 (Growth comparison)

For every integer $n \geq 1$ ,

\lim_{x\to+\infty} \frac{\eu^x}{x^n} = +\infty, \qquad \lim_{x\to-\infty} x^n \eu^x = 0 .

In words: the exponential beats every power of $x$ .

Proof. For $n = 1$ : applying $\eu^t \geq 1 + t \geq t$ at $t = x/2$ ,

\frac{\eu^x}{x} = \frac{\left(\eu^{x/2}\right)^2}{x} \geq \frac{(x/2)^2}{x} = \frac{x}{4} \xrightarrow[x\to+\infty]{} +\infty .

For general $n$ , write

\frac{\eu^x}{x^n} = \frac{1}{n^n}\left(\frac{\eu^{x/n}}{x/n}\right)^{n} \xrightarrow[x\to+\infty]{} +\infty

by the case $n=1$ and composition. The limit at $-\infty$ follows by the substitution $x \mapsto -x$ : $\abs{x^n \eu^x} = \frac{\abs{x}^n}{\eu^{\abs x}} \to 0$ . ∎

The exponential (blue) eventually beats every power: it overtakes x2 (red) early on and x3 (orange) at x 4.54. — The exponential (blue) eventually beats every power: it overtakes $x^2$ (red) early on and $x^3$ (orange) at $x \approx 4.54$ .

23.2 The natural logarithm

Definition 23.5 (Natural logarithm)

The exponential is continuous and strictly increasing from $\R$ onto $\intoo{0}{+\infty}$ ; by the bijection theorem (Theorem 21.15), for every $y > 0$ the equation $\eu^x = y$ has a unique solution. This solution is the natural logarithm of $y$ , written $\ln y$ . Thus

\text{for } x \in \R,\ y > 0: \qquad y = \eu^x \iff x = \ln y .

In particular $\ln 1 = 0$ , $\ln \eu = 1$ , and $\eu^{\ln y} = y$ , $\ln(\eu^x) = x$ .

The curves of and are mirror images of each other in the line y = x: the point (x, x) reflects to ( x, x). — The curves of $\exp$ and $\ln$ are mirror images of each other in the line $y = x$ : the point $(x, \eu^x)$ reflects to $(\eu^x, x)$ .

Proposition 23.6 (Algebraic properties)

For all $a, b > 0$ and $n \in \Z$ :

\ln(ab) = \ln a + \ln b, \quad \ln\frac{1}{a} = -\ln a, \quad \ln\frac{a}{b} = \ln a - \ln b, \quad \ln(a^n) = n \ln a, \quad \ln\sqrt{a} = \tfrac12 \ln a .

Proof. $\eu^{\ln a + \ln b} = \eu^{\ln a}\eu^{\ln b} = ab$ , and taking $\ln$ of both sides gives the first identity. The others follow by the same mechanism from the corresponding identities of Proposition 23.2. ∎

Proposition 23.7 (Analytic properties)

The function $\ln$ is differentiable on $\intoo{0}{+\infty}$ with

(\ln x)' = \frac{1}{x},

strictly increasing, concave, and $\lim\limits_{x\to0^+} \ln x = -\infty$ , $\lim\limits_{x\to+\infty} \ln x = +\infty$ . Moreover, for every integer $n \geq 1$ ,

\lim_{x\to+\infty} \frac{\ln x}{x^{1/n}} = 0, \qquad \lim_{x\to0^+} x \ln x = 0 .

Proof. Differentiability of the inverse function is admitted at this level; granting it, differentiate the identity $\eu^{\ln x} = x$ by the chain rule: $(\ln x)'\,\eu^{\ln x} = 1$ , so $(\ln x)' = \frac{1}{x} > 0$ , whence strict increase, and $(\ln x)'' = -\frac1{x^2} < 0$ , whence concavity. The limits at $0^+$ and $+\infty$ mirror those of $\exp$ through the bijection. For the growth comparison, substitute $x = \eu^{t}$ : $\frac{\ln x}{x} = \frac{t}{\eu^t} \to 0$ as $t \to +\infty$ by Theorem 23.4, and similarly for the other limits with $x = \eu^{-t}$ , giving $x\ln x = -t\eu^{-t} \to 0$ . ∎

Method 23.8 (Solving equations with $\exp$ and $\ln$ )

Both functions are strictly increasing, so they can be applied to (or removed from) both sides of an equation or inequality without changing its direction:

\eu^{u} = \eu^{v} \iff u = v, \qquad \eu^{u} \leq \eu^{v} \iff u \leq v,

and likewise for $\ln$ on positive quantities. Always check domains first ( $\ln$ requires positive arguments). For equations like $a^x = b$ with $a>0$ , $a \neq 1$ , rewrite $a^x = \eu^{x\ln a}$ and solve $x = \frac{\ln b}{\ln a}$ .

Example 23.9

A radioactive sample decays following $N(t) = N_0\,\eu^{-\lambda t}$ . Its half-life $T$ satisfies $N(T) = N_0/2$ , i.e. $\eu^{-\lambda T} = \frac12$ , so $T = \frac{\ln 2}{\lambda}$ : the half-life does not depend on the initial quantity.

23.3 Exercises

Exercise 23.1 ★

Simplify $\dfrac{\eu^{3x}\,\eu^{-x+1}}{\eu^{x}}$ and $\ln\!\left(\dfrac{\eu^2\sqrt{\eu}}{\eu^{-3}}\right)$ .

Solution

Solution of Exercise 23.1.

$\dfrac{\eu^{3x}\,\eu^{-x+1}}{\eu^{x}} = \eu^{3x - x + 1 - x} = \eu^{x+1}$ .

$\ln\!\left(\dfrac{\eu^2\sqrt{\eu}}{\eu^{-3}}\right) = 2 + \tfrac12 + 3 = \tfrac{11}{2}$ .

Exercise 23.2 ★

Solve in $\R$ :

\text{(a) } \eu^{2x} - 3\eu^{x} + 2 = 0; \qquad \text{(b) } \ln(x - 1) + \ln(x + 2) = \ln 4 .

Solution

Solution of Exercise 23.2.

(a) Set $X = \eu^x > 0$ : $X^2 - 3X + 2 = 0$ gives $X = 1$ or $X = 2$ , both positive, so $x = 0$ or $x = \ln 2$ .

(b) Domain: $x - 1 > 0$ and $x + 2 > 0$ , so $x > 1$ . The equation becomes $\ln\bigl((x-1)(x+2)\bigr) = \ln 4$ , hence $(x-1)(x+2) = 4$ , i.e. $x^2 + x - 6 = 0$ , so $x = 2$ or $x = -3$ . Only $x = 2$ is in the domain: $S = \{2\}$ .

Exercise 23.3 ★

Compute the limits:

\lim_{x\to+\infty} \frac{\eu^x - x^2}{\eu^x + 1}, \qquad \lim_{x\to+\infty} \frac{\ln x}{\sqrt x}, \qquad \lim_{x\to0^+} x^2 \ln x, \qquad \lim_{x\to+\infty} \bigl(x - \ln x\bigr).

Solution

Solution of Exercise 23.3.

Divide by $\eu^x$ : $\dfrac{1 - x^2\eu^{-x}}{1 + \eu^{-x}} \to \dfrac{1-0}{1+0} = 1$ , using $x^2 \eu^{-x} \to 0$ (Theorem 23.4).

$\dfrac{\ln x}{\sqrt x} \to 0$ by Proposition 23.7 (case $n = 2$ ).

$x^2 \ln x = x \cdot (x \ln x) \to 0 \times 0 = 0$ .

$x - \ln x = x\left(1 - \frac{\ln x}{x}\right) \to +\infty$ since $\frac{\ln x}{x} \to 0$ .

Exercise 23.4 ★★

Study the function $f(x) = x\,\eu^{-x}$ on $\R$ : variations, limits, extremum; show that its curve has an inflection point and give its coordinates.

Solution

Solution of Exercise 23.4.

$f'(x) = \eu^{-x} - x\eu^{-x} = (1 - x)\eu^{-x}$ , of the sign of $1 - x$ : $f$ increases on $\intoc{-\infty}{1}$ , decreases on $\intco{1}{+\infty}$ , with a global maximum $f(1) = \eu^{-1}$ .

Limits: as $x \to +\infty$ , $f(x) = \frac{x}{\eu^x} \to 0$ (Theorem 23.4); as $x \to -\infty$ , $x \to -\infty$ and $\eu^{-x} \to +\infty$ , so $f(x) \to -\infty$ .

$f''(x) = -\eu^{-x} - (1-x)\eu^{-x} = (x - 2)\eu^{-x}$ , which changes sign at $x = 2$ : inflection point at $\bigl(2,\, 2\eu^{-2}\bigr)$ .

Exercise 23.5 ★★

Show that for all $x > -1$ , $\ln(1 + x) \leq x$ , with equality only at $x = 0$ . Deduce that for all $n \geq 1$ ,

\left(1 + \frac{1}{n}\right)^n \leq \eu \leq \left(1 - \frac{1}{n+1}\right)^{-(n+1)} .

Solution

Solution of Exercise 23.5.

Let $g(x) = x - \ln(1+x)$ on $\intoo{-1}{+\infty}$ . Then $g'(x) = 1 - \frac{1}{1+x} = \frac{x}{1+x}$ , negative on $\intoo{-1}{0}$ and positive on $\intoo{0}{+\infty}$ : $g$ attains its minimum $g(0) = 0$ , so $g \geq 0$ with equality only at $0$ . Hence $\ln(1+x) \leq x$ .

Apply this with $x = \frac1n$ : $n\ln\left(1 + \frac1n\right) \leq 1$ , so $\left(1+\frac1n\right)^n = \eu^{\,n\ln(1+1/n)} \leq \eu$ . Apply it with $x = -\frac{1}{n+1} > -1$ : $\ln\left(1 - \frac{1}{n+1}\right) \leq -\frac{1}{n+1}$ , so $-(n+1)\ln\left(1 - \frac{1}{n+1}\right) \geq 1$ and $\left(1 - \frac{1}{n+1}\right)^{-(n+1)} \geq \eu$ .

Exercise 23.6 ★★

A capital $C_0$ is invested at an annual rate of $3\%$ , interest compounded each year.

Express the capital $C_n$ after $n$ years.
After how many years does the capital double? Give the exact answer using $\ln$ , then a numerical value.
Compare with the approximation “ $70$ divided by the rate in percent” used by bankers.

Solution

Solution of Exercise 23.6.

1. Each year multiplies the capital by $1.03$ : $C_n = C_0 (1.03)^n$ .

2. $C_n \geq 2C_0 \iff (1.03)^n \geq 2 \iff n \ln 1.03 \geq \ln 2 \iff n \geq \frac{\ln 2}{\ln 1.03} \approx 23.45$ : the capital doubles after $24$ years.

3. $\frac{70}{3} \approx 23.3$ , close to the exact $\frac{\ln 2}{\ln 1.03}$ . The rule works because $\ln(1 + r) \approx r$ for small $r$ , so $\frac{\ln 2}{\ln(1+r)} \approx \frac{0.693}{r} \approx \frac{70}{100r}$ .

Exercise 23.7 ★★

Solve the inequality $\eu^{2x} - \eu^{x+1} > 0$ , then the inequality $\ln(x^2 - 1) \leq \ln(x + 5)$ .

Solution

Solution of Exercise 23.7.

$\eu^{2x} > \eu^{x+1} \iff 2x > x + 1 \iff x > 1$ (the exponential is strictly increasing): $S = \intoo{1}{+\infty}$ .

Domain of the second inequality: $x^2 - 1 > 0$ and $x + 5 > 0$ , i.e. $x \in \intoo{-5}{-1} \cup \intoo{1}{+\infty}$ . On this domain, $\ln$ being strictly increasing,

\ln(x^2-1) \leq \ln(x+5) \iff x^2 - 1 \leq x + 5 \iff x^2 - x - 6 \leq 0 \iff x \in \intcc{-2}{3}.

Intersecting with the domain: $S = \intco{-2}{-1} \cup \intoc{1}{3}$ .

Exercise 23.8 ★★★

Let $f(x) = \dfrac{\eu^x}{x}$ for $x > 0$ .

Study the variations of $f$ on $\intoo{0}{+\infty}$ and give its minimum.
For which values of $k$ does the equation $\eu^x = kx$ have $0$ , $1$ or $2$ solutions in $\intoo{0}{+\infty}$ ?

Solution

Solution of Exercise 23.8.

1. $f'(x) = \dfrac{\eu^x x - \eu^x}{x^2} = \dfrac{(x-1)\eu^x}{x^2}$ , negative on $\intoo{0}{1}$ , positive on $\intoo{1}{+\infty}$ : minimum $f(1) = \eu$ . Limits: $f \to +\infty$ at $0^+$ (numerator $\to 1$ , denominator $\to 0^+$ ) and at $+\infty$ (Theorem 23.4).

2. For $x > 0$ , $\eu^x = kx \iff f(x) = k$ . From the variation table ( $+\infty \searrow \eu \nearrow +\infty$ , continuous and strictly monotonic on each side): no solution for $k < \eu$ ; exactly one ( $x = 1$ ) for $k = \eu$ ; exactly two for $k > \eu$ (one in $\intoo{0}{1}$ , one in $\intoo{1}{+\infty}$ , by the bijection theorem on each interval).

Exercise 23.9 ★★★

For $n \geq 1$ , let $u_n = \left(1 + \frac1n\right)^n$ .

Using Exercise 23.5, show that $(u_n)$ is bounded above by $\eu$ .
Show that $\ln u_n = n \ln\left(1 + \frac1n\right) \to 1$ , and deduce that $u_n \to \eu$ . (Hint: recognize a difference quotient of $\ln$ at $1$ .)

Solution

Solution of Exercise 23.9.

1. Direct from the first inequality of Exercise 23.5: $u_n = \eu^{\,n\ln(1+1/n)} \leq \eu^1 = \eu$ .

2. Write

\ln u_n = n \ln\!\left(1 + \frac1n\right) = \frac{\ln\!\left(1 + \frac1n\right) - \ln 1}{\frac1n},

a difference quotient of $\ln$ at the point $1$ with increment $h = \frac1n \to 0$ . Since $\ln$ is differentiable at $1$ with derivative $1$ , $\ln u_n \to 1$ . By continuity of $\exp$ (Proposition 21.12), $u_n = \eu^{\ln u_n} \to \eu^1 = \eu$ .

23.4 Problem: The logarithm tames the world

Problem 23.1

Weekend problem — doubling times and the rule of 72, the scales of earthquakes, acids and pianos, and the constant $\eu$ leaving fingerprints everywhere

Whatever grows by a fixed percentage grows exponentially — savings, bacteria, epidemics — and whatever spans too many powers of ten to grasp — earthquake energies, acidities, sound intensities — is tamed by a logarithm. This problem computes doubling times and unmasks the bankers’ rule of 72, reads the world’s logarithmic scales, and collects the fingerprints that the number $\eu$ leaves at every crime scene (Proposition 23.2, Theorem 23.4, Method 23.8).

Part I — Fluency.

Solve: $\eu^{2x} = 5$ ; $\ln(3x - 1) = 2$ ; $\eu^{2x} - 3\eu^x + 2 = 0$ (a quadratic in disguise).
Simplify: $\dfrac{\ln 8}{\ln 2}$ ; $\ln\!\left(\eu^3 \sqrt{\eu}\right)$ ; $\eu^{\ln 5 - \ln 2}$ .
Prove the mother inequality: $\eu^x \geq 1 + x$ for all real $x$ (study $f(x) = \eu^x - 1 - x$ ), and deduce $\ln(1 + u) \leq u$ for $u > -1$ .
Compute $\lim_{x \to +\infty} x^2 \eu^{-x}$ , $\lim_{x \to +\infty} \frac{\ln x}{x}$ (Theorem 23.4), and $\lim_{x \to 0} \frac{\eu^x - 1}{x}$ (recognize a derivative).
Study $f(x) = x \ln x$ on $\intoo{0}{+\infty}$ : variations, minimum, and the limit at $0^+$ (admitted: $x \ln x \to 0$ ). This little function measures information and entropy across the university volumes.

Part II — Doubling times and the rule of 72.

Savings grow at $3\,\%$ per year. Solve $1.03^n = 2$ : how long to double the capital?
Bankers estimate doubling time as $\frac{72}{\text{rate in }\%}$ . Test the rule at $3\,\%$ , $6\,\%$ and $9\,\%$ against the exact $\frac{\ln 2}{\ln(1 + r)}$ , then explain it: for small $r$ , $\ln(1 + r) \approx r$ (question 3’s inequality is half of the story), so the exact constant is $100 \ln 2 \approx 69.3$ — why do bankers prefer $72$ ?
A bacterium divides every $20$ minutes: $p(t) = 2^{t/20}$ ( $t$ in minutes). Rewrite it as $\eu^{\lambda t}$ , then compute $p$ after $24$ hours. The answer (more than $10^{21}$ ) proves what about the model — and what stops real colonies?
Caffeine leaves the body with a half-life of about $5$ hours. After a $100$ mg coffee at 15:00, how much remains at 23:00? At what time does it drop below $10$ mg? (Insomnia has a logarithm.)
One euro at $100\,\%$ annual interest: compute the year-end capital under yearly, monthly and daily compounding, and give the ceiling that Exercise 23.9 proved unbreakable. Which famous constant is the limit of pure greed?
Why do scientists plot $\ln(\text{cases})$ against time during an epidemic’s early phase? What does a straight line on that plot reveal, and what does its slope measure?

Part III — The world’s logarithmic scales. (Write $\log_{10} x = \frac{\ln x}{\ln 10}$ .)

Sound level in decibels: $L = 10 \log_{10}(I/I_0)$ . A conversation measures $60$ dB, a rock concert $120$ dB: by what factor do the sound intensities differ?
Earthquake magnitudes rise by $1$ when the seismic amplitude is multiplied by $10$ , and the released energy scales like amplitude $^{3/2}$ . Compare magnitude- $5$ and magnitude- $7$ quakes: amplitude ratio, then energy ratio.
Chemistry: $\text{pH} = -\log_{10}[\mathrm{H^+}]$ . Lemon juice has pH $2$ , milk pH $6.5$ : what is the ratio of their acid concentrations?
Music: each octave doubles the frequency. From the piano’s lowest A ( $27.5$ Hz) to its highest C ( $4\,186$ Hz), how many octaves does the keyboard span? And why do our senses — hearing, sight, quake-feeling — prefer logarithmic scales? (One sentence.)
The slide rule, the engineers’ calculator for $350$ years: two sticks graduated so that the length to the mark $x$ is proportional to $\ln x$ . Explain how sliding one stick along the other multiplies numbers, and name the identity of Proposition 23.6 doing the work.

Part IV — $\eu$ ’s fingerprints.

From Exercise 23.8: the equation $\eu^x = kx$ ( $x > 0$ ) has $0$ , $1$ or $2$ solutions according to the position of $k$ relative to a threshold. Restate the result — and verify the geometric fact behind it: the line $y = \eu x$ is exactly the tangent to the exponential through the origin.
The near-miss constant: with $n$ lottery tickets of winning probability $\frac1n$ each, $\P(\text{no win}) = \left(1 - \frac1n\right)^n$ . Compute its limit via $n \ln\!\left(1 - \frac1n\right)$ (use the standard limit $\frac{\ln(1 + u)}{u} \to 1$ ), and reconcile with the mysterious $0.368$ of Problem 19.1.
The $37\,\%$ rule: to choose the best of $n$ candidates interviewed in random order (no going back), the optimal strategy rejects the first $\frac{n}{\eu} \approx 37\,\%$ and then takes the first candidate better than all so far — succeeding with probability about $\frac1\eu$ . Where do the two $\frac1\eu$ ’s of questions 18 and 19 come from — state the common mechanism (many independent small-chance events), and compute $\frac1\eu$ to three decimals.
Finale — $\eu$ ’s portrait: the ceiling of compounding (question 10); the base whose tangent at $0$ has slope exactly $1$ (questions 3 and 4); the growth no polynomial catches (question 4); the constant of near-misses and of optimal stopping (questions 18–19); and its inverse $\ln$ , which turns products into sums (question 16) and decades into inches (Part III). One sentence each.

Solution

Solution of Problem 23.1.

1. $x = \frac{\ln 5}{2}$ . $3x - 1 = \eu^2$ : $x = \frac{\eu^2 + 1}{3}$ . With $u = \eu^x$ : $u^2 - 3u + 2 = (u - 1)(u - 2) = 0$ : $x = 0$ or $x = \ln 2$ .

2. $\frac{3\ln 2}{\ln 2} = 3$ ; $3 + \frac12 = \frac72$ ; $\frac52$ .

3. $f'(x) = \eu^x - 1$ : negative before $0$ , positive after: minimum $f(0) = 0$ , so $f \geq 0$ everywhere: $\eu^x \geq 1 + x$ . Substituting $x = \ln(1 + u)$ : $1 + u \geq 1 + \ln(1 + u)$ , i.e. $\ln(1 + u) \leq u$ .

4. $x^2\eu^{-x} = \frac{x^2}{\eu^x} \to 0$ and $\frac{\ln x}{x} \to 0$ : exponentials crush powers, powers crush logarithms. And $\frac{\eu^x - 1}{x} = \frac{\eu^x - \eu^0}{x - 0} \to \exp'(0) = 1$ .

5. $f'(x) = \ln x + 1$ : zero at $\eu^{-1}$ , negative before, positive after: minimum $f\!\left(\frac1\eu\right) = -\frac1\eu$ ; and $x \ln x \to 0$ at $0^+$ : the curve leaves the origin, dips to $-\frac1\eu$ , and climbs away.

6. $n = \frac{\ln 2}{\ln 1.03} \approx 23.4$ years.

7. Exact: $23.4$ , $11.9$ , $8.0$ years; rule of 72: $24$ , $12$ , $8$ . Since $\ln(1+r) \approx r$ , $\frac{\ln 2}{\ln(1+r)} \approx \frac{0.693}{r}$ , i.e. $\frac{69.3}{\text{rate in }\%}$ ; bankers round up to $72$ because it divides beautifully by $2, 3, 4, 6, 8, 9, 12$ — mental arithmetic beats a decimal of accuracy.

8. $2^{t/20} = \eu^{t \ln 2 / 20}$ : $\lambda = \frac{\ln 2}{20}$ per minute. After $24 \times 60 = 1440$ minutes: $2^{72} \approx 4.7 \times 10^{21}$ bacteria — more than the grains of sand on Earth, from one cell in one day. The model is honest only while food and space last: real growth bends into the logistic S-curve (the epidemiologist’s curve of Problem 22.1).

9. At 23:00 ( $8$ hours): $100 \times 2^{-8/5} \approx 33$ mg. Below $10$ mg: $2^{-t/5} < 0.1$ gives $t > 5\,\frac{\ln 10}{\ln 2} \approx 16.6$ h: around 7:40 the next morning — the espresso at three has a long tail.

10. Yearly: $2$ . Monthly: $\left(1 + \frac{1}{12}\right)^{12} \approx 2.613$ . Daily: $\approx 2.715$ . The ceiling: $\eu = 2.71828\ldots$ (Exercise 23.9) — compounding continuously, greed converges.

11. If cases grow exponentially, $c(t) = c_0 \eu^{\lambda t}$ , then $\ln c(t) = \ln c_0 + \lambda t$ : a straight line of slope $\lambda$ — the growth rate. A straight stretch on the log plot is the exponential phase, and its steepness is the epidemic’s tempo.

12. $120 - 60 = 60$ dB means $10\log_{10}(I_2/I_1) = 60$ : the concert is $10^6$ — a million times — more intense than the conversation.

13. Amplitude: $10^{7-5} = 100$ . Energy: $100^{3/2} = 1000$ : two magnitude points hide three orders of magnitude in energy.

14. $10^{6.5 - 2} = 10^{4.5} \approx 32\,000$ : lemon juice is thirty thousand times more acidic than milk — pH compresses chemistry’s chasms into a pocket scale.

15. $\log_2\!\frac{4186}{27.5} = \frac{\ln(4186/27.5)}{\ln 2} \approx 7.25$ : a piano spans just over seven octaves. Senses respond to ratios of stimuli — doubling the intensity feels like one step, whatever the starting level — so perception is built on a logarithmic scale, and so are the units we invented for it.

16. Placing the stick for $a$ end-to-end with the stick position for $b$ adds the lengths $\ln a + \ln b$ , and the graduation sitting at that total length reads $\eu^{\ln a + \ln b} = ab$ : the slide rule computes products by adding logarithms — $\ln(ab) = \ln a + \ln b$ (Proposition 23.6) carved in boxwood.

17. The minimum of $\frac{\eu^x}{x}$ on $\intoo{0}{+\infty}$ is $\eu$ , at $x = 1$ : no solution for $k < \eu$ , exactly one for $k = \eu$ , two for $k > \eu$ . Tangent check: at $a = 1$ the tangent to $\eu^x$ is $y = \eu + \eu(x - 1) = \eu x$ : it passes through the origin — the threshold line, grazing the curve at $(1, \eu)$ .

18. $n \ln\!\left(1 - \frac1n\right) = \frac{\ln(1 - 1/n)}{-1/n} \times (-1) \to -1$ , so $\left(1 - \frac1n\right)^n \to \eu^{-1} \approx 0.368$ : the $0.999^{1000} \approx 0.368$ of Problem 19.1, explained — the lottery’s near-miss constant is $\frac1\eu$ .

19. Both are the limit shape of “many independent events, each individually unlikely”: the chance that none of $n$ chances of size $\frac1n$ fires tends to $\eu^{-1}$ , and the secretary rule tunes its rejection window so that success concentrates at that same constant. $\frac1\eu \approx 0.368$ .

20. The ceiling of compounding; the unique base whose tangent at $0$ has slope $1$ (which is why calculus loves it); the growth that outruns every power; the constant where near-misses and optimal stopping settle; and the logarithm — multiplication become addition, the world’s wildest ranges folded onto a ruler.