---
title: "Exponential and Logarithm"
book: "High School Mathematics"
subject: math
language: en
chapter: 23
exercises: 9
source: https://one-course.com/books/math/2/en/chapter/23-exponential-and-logarithm
---

# Chapter 23 — Exponential and Logarithm

The [exponential function](#thm-g12-exp-existence) is the unique [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) equal to its own [derivative](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative) and taking the value $1$ at $0$. It converts sums into products; its inverse, the [natural logarithm](#def-g12-exp-ln), 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](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative) [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-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](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-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](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-sequence) of [Exercise 20.10](https://one-course.com/books/math/2/en/chapter/20-sequences#exo-g12-seq-10), or as the inverse of the [logarithm](#def-g12-exp-ln) constructed by integration in [Chapter 25](https://one-course.com/books/math/2/en/chapter/25-integration#ch-g12-integ)). ∎

**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](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) 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](#thm-g12-exp-existence)), so $\eu^x > 0$. ∎

**Proposition 23.3 (Variations and limits).**

The exponential is strictly [increasing](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-monotonic), [convex](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-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](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-convex). By [convexity](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-convex), $\eu^x \geq 1 + x$ ([tangent](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-tangent) at $0$, see [Example 22.15](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#ex-g12-deriv-ineq)), 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.](https://one-course.com/images/onecourse/chapters/math-2/g12-exp/fig-4d7aa83a2aa6.svg)

*The exponential: strictly [increasing](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-monotonic), [convex](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-convex), with $\lim_{-\infty} \eu^x = 0$. It lies above its [tangent](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-tangent) at $0$: $\eu^x \geq 1 + x$.*

**Theorem 23.4 (Growth comparison).**

For every [integer](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-sets) $n \geq 1$,

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

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.](https://one-course.com/images/onecourse/chapters/math-2/g12-exp/fig-2bf938e48637.svg)

*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](https://one-course.com/books/math/2/en/chapter/21-limits-and-continuity#def-g12-limcont-continuity) and strictly [increasing](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-monotonic) from $\R$ onto $\intoo{0}{+\infty}$; by the bijection theorem ([Theorem 21.15](https://one-course.com/books/math/2/en/chapter/21-limits-and-continuity#thm-g12-limcont-bijection)), for every $y > 0$ the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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).](https://one-course.com/images/onecourse/chapters/math-2/g12-exp/fig-a625fce5956d.svg)

*The curves of $\exp$ and $\ln$ are mirror [images](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) 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](#prop-g12-exp-funceq). ∎

**Proposition 23.7 (Analytic properties).**

The [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) $\ln$ is [differentiable](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative) on $\intoo{0}{+\infty}$ with

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

strictly [increasing](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-monotonic), [concave](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-convex), and $\lim\limits_{x\to0^+} \ln x = -\infty$, $\lim\limits_{x\to+\infty} \ln x = +\infty$. Moreover, for every [integer](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-sets) $n \geq 1$,

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

**Proof.** Differentiability of the inverse [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-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](#thm-g12-exp-growth), substitute $x = \eu^{t}$: $\frac{\ln x}{x} = \frac{t}{\eu^t} \to 0$ as $t \to +\infty$ by [Theorem 23.4](#thm-g12-exp-growth), 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⁡\expexp and ln⁡\lnln).**

Both [functions](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) are strictly [increasing](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-monotonic), so they can be applied to (or removed from) both sides of an [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) first* ($\ln$ requires positive arguments). For [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) 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](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-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 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 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](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function): $x - 1 > 0$ and $x + 2 > 0$, so $x > 1$. The [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function): $S = \{2\}$.

**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 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](#thm-g12-exp-growth)).

$\dfrac{\ln x}{\sqrt x} \to 0$ by [Proposition 23.7](#prop-g12-exp-lnanalytic) (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](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) $f(x) = x\,\eu^{-x}$ on $\R$: variations, limits, [extremum](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-extremum); show that its curve has an [inflection point](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-inflection) and give its [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system).

**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](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) $f(1) = \eu^{-1}$.

Limits: as $x \to +\infty$, $f(x) = \frac{x}{\eu^x} \to 0$ ([Theorem 23.4](#thm-g12-exp-growth)); 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](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-inflection) 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 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](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) $g(0) = 0$, 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.

1. Express the capital $C_n$ after $n$ years.
2. After how many years does the capital double? Give the exact answer using $\ln$ , then a numerical value.
3. Compare with the [approximation](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-approx) “ $70$ divided by the rate in percent” used by bankers.

**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 of Exercise 23.7.**

$\eu^{2x} > \eu^{x+1} \iff 2x > x + 1 \iff x > 1$ (the exponential is strictly [increasing](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-monotonic)): $S = \intoo{1}{+\infty}$.

[Domain](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) 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](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function), $\ln$ being strictly [increasing](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-monotonic),

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

Intersecting with the [domain](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function): $S = \intco{-2}{-1} \cup \intoc{1}{3}$.

**Exercise 23.8 ★★★.**

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

1. Study the variations of $f$ on $\intoo{0}{+\infty}$ and give its [minimum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) .
2. For which values of $k$ does the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $\eu^x = kx$ have $0$ , $1$ or $2$ solutions in $\intoo{0}{+\infty}$ ?

**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](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) $f(1) = \eu$. Limits: $f \to +\infty$ at $0^+$ (numerator $\to 1$, denominator $\to 0^+$) and at $+\infty$ ([Theorem 23.4](#thm-g12-exp-growth)).

*2.* For $x > 0$, $\eu^x = kx \iff f(x) = k$. From the [variation table](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-table) ($+\infty \searrow \eu \nearrow +\infty$, [continuous](https://one-course.com/books/math/2/en/chapter/21-limits-and-continuity#def-g12-limcont-continuity) and strictly [monotonic](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-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](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval)).

**Exercise 23.9 ★★★.**

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

1. Using [Exercise 23.5](#exo-g12-exp-5) , show that $(u_n)$ is [bounded above](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-bounded) by $\eu$ .
2. 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 of Exercise 23.9.**

*1.* Direct from the first inequality of [Exercise 23.5](#exo-g12-exp-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](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative) at $1$ with [derivative](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative) $1$, $\ln u_n \to 1$. By [continuity](https://one-course.com/books/math/2/en/chapter/21-limits-and-continuity#def-g12-limcont-continuity) of $\exp$ ([Proposition 21.12](https://one-course.com/books/math/2/en/chapter/21-limits-and-continuity#prop-g12-limcont-seqcont)), $u_n = \eu^{\ln u_n} \to \eu^1 = \eu$.

## 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](#def-g12-exp-ln). 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](#prop-g12-exp-funceq), [Theorem 23.4](#thm-g12-exp-growth), [Method 23.8](#met-g12-exp-equations)).

**Part I — Fluency.**

1. Solve: $\eu^{2x} = 5$ ; $\ln(3x - 1) = 2$ ; $\eu^{2x} - 3\eu^x + 2 = 0$ (a quadratic in disguise).
2. Simplify: $\dfrac{\ln 8}{\ln 2}$ ; $\ln\!\left(\eu^3 \sqrt{\eu}\right)$ ; $\eu^{\ln 5 - \ln 2}$ .
3. 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 $u > -1$ .
4. Compute $\lim_{x \to +\infty} x^2 \eu^{-x}$ , $\lim_{x \to +\infty} \frac{\ln x}{x}$ ( [Theorem 23.4](#thm-g12-exp-growth) ), and $\lim_{x \to 0} \frac{\eu^x - 1}{x}$ (recognize a [derivative](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative) ).
5. Study $f(x) = x \ln x$ on $\intoo{0}{+\infty}$ : variations, [minimum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) , and the limit at $0^+$ (admitted: $x \ln x \to 0$ ). This little [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) measures information and entropy across the university volumes.

**Part II — Doubling times and the rule of 72.**

6. Savings grow at $3\,\%$ per year. Solve $1.03^n = 2$ : how long to double the capital?
7. 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 $72$ ?
8. 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?
9. 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](#def-g12-exp-ln) .)
10. 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](#exo-g12-exp-9) proved unbreakable. Which famous constant is the limit of pure greed?
11. 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](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) measure?

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

12. 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?
13. 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.
14. Chemistry: $\text{pH} = -\log_{10}[\mathrm{H^+}]$ . Lemon juice has pH $2$ , milk pH $6.5$ : what is the ratio of their acid concentrations?
15. Music: each octave doubles the [frequency](https://one-course.com/books/math/2/en/chapter/8-descriptive-statistics#def-g10-stats-series) . 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.)
16. 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](#prop-g12-exp-lnalg) doing the work.

**Part IV — $\eu$’s fingerprints.**

17. From [Exercise 23.8](#exo-g12-exp-8) : the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-tangent) to the exponential through the origin.
18. The near-miss constant: with $n$ lottery tickets of winning [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $\frac1n$ 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](https://one-course.com/books/math/2/en/chapter/19-the-binomial-distribution#pb-g11-binom-1) .
19. 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](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) 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](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#def-g11-prob-model) ), and compute $\frac1\eu$ to three decimals.
20. Finale — $\eu$ ’s portrait: the ceiling of compounding (question 10); the base whose [tangent](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-tangent) at $0$ has [slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) 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 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](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) $f(0) = 0$, 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](#def-g12-exp-ln). 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](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) $f\!\left(\frac1\eu\right) = -\frac1\eu$; 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](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#pb-g12-deriv-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](#exo-g12-exp-9)) — compounding [continuously](https://one-course.com/books/math/2/en/chapter/21-limits-and-continuity#def-g12-limcont-continuity), greed [converges](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-limit).

**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](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) $\lambda$ — 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](#def-g12-exp-ln) — $\ln(ab) = \ln a + \ln b$ ([Proposition 23.6](#prop-g12-exp-lnalg)) carved in boxwood.

**17.** The [minimum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) 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](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-tangent) check: at $a = 1$ the [tangent](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-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](https://one-course.com/books/math/2/en/chapter/19-the-binomial-distribution#pb-g11-binom-1), explained — the lottery’s near-miss constant is $\frac1\eu$.

**19.** Both are the limit shape of “many independent [events](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#def-g11-prob-model), 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](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-tangent) at $0$ has [slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) $1$ (which is why calculus loves it); the growth that outruns every power; the constant where near-misses and optimal stopping settle; and the [logarithm](#def-g12-exp-ln) — multiplication become addition, the world’s wildest ranges folded onto a ruler.