8.5 The Exponential, the Logarithm, and Real Powers

Chapter 1 gave every non-negative real number its $n$ -th roots (Lemma 1.5.1). It gave every positive real number its rational powers (Definition 1.5.2), so a power with a rational exponent such as $2^{3/2} = (\sqrt 2)^3$ already has a meaning. A power such as $2^{\sqrt 2}$ does not, and no root will give it one. This section borrows three facts about a single function, the exponential, and derives everything else from them: the logarithm, real powers and their laws, and how fast the exponential grows. Its main theorem shows that two of the three facts already determine the function completely.

8.5.1. Powers that roots cannot reach

A power $a^{m/n}$ with $a > 0$ is a power of an $n$ -th root, $(\sqrt[n]{a})^m$ . The exponent $\sqrt 2$ is not of the form $m/n$ , because $\sqrt 2$ is irrational, as Section 8.3 showed from Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 3.3.1. So $2^{\sqrt 2}$ is not a root of anything we know.

The naive answer, “the limit of $2^q$ as rational $q$ approach $\sqrt 2$ ”, needs limits of functions, which this volume does not have. Our route goes through one function, the exponential, and one rule, $a^x \mathrel{:=} \exp(x \log a)$ . Constructing the exponential needs power series, so we borrow it: three facts, and everything else is proved.

8.5.2. What this volume borrows about the exponential

The exponential should turn sums into products, and near $0$ it should look like $1 + x$ . Integration, Series, and Function Spaces, Chapter 7 constructs it as a power series and proves the facts below. Its $\mathbb{R}$ is a complete ordered field, so by Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 6.3.1 it is our $\mathbb{R}$ , up to the one order-preserving field isomorphism, and the function carries over along it.

Axiom 8.5.1 (What This Volume Borrows about the Exponential)

There is a function $\exp \colon \mathbb{R} \to \mathbb{R}$ with the following properties. They are assumed here and proved in Integration, Series, and Function Spaces, Chapter 7.

(E1) $\exp(x + y) = \exp x \cdot \exp y$ for all $x, y \in \mathbb{R}$ .
(E2) $\exp x \ge 1 + x$ for every $x \in \mathbb{R}$ .
(E3) For every $y \in \mathbb{R}$ with $y > 0$ there is $x \in \mathbb{R}$ with $\exp x = y$ .

In words: the addition rule (E1) turns addition of arguments into multiplication of values. The tangent-line bound (E2) puts the graph of $\exp$ on or above the line $y = 1 + x$ . The value property (E3) says that every positive real number is a value of $\exp$ .

Chapter 3 borrowed cosine and sine in the same way (Axiom 3.5.2). Nothing in Sections 8.1–8.4, and nothing earlier in this section, uses these three facts. From here on we use these three facts about $\exp$ and no others, and each proof that uses one names it. The addition rule and the tangent-line bound carry most of the weight. The value property is what makes the logarithm defined on all of $(0, \infty)$ .

Neither of the first two facts follows from the other. The constant function $1$ satisfies the addition rule, since $1 = 1 \cdot 1$ , but not the tangent-line bound: at $x = 1$ it would need $1 \ge 2$ . The function $x \mapsto 1 + x$ satisfies the tangent-line bound with equality, but not the addition rule: at $x = y = 1$ it gives $3$ , while $(1 + 1)(1 + 1) = 4$ .

Warning. The three facts say nothing about continuity or derivatives. That $\exp$ is continuous, and equal to its own derivative, is true. But it is not among the three facts, and no proof in this volume may quote it. Nor is $e^x$ a notation yet. Until real powers are defined and shown to agree with $\exp$ , we write $\exp x$ , never $e^x$ .

8.5.3. First consequences

Negative integer exponents appear as soon as we divide. For a real $u \ne 0$ and $n \in \mathbb{Z}_{>0}$ , the power $u^{-n} = (u^{-1})^n = (1/u)^n$ is the integer power of Chapter 1 (Definition 1.5.1). By Proposition 1.5.3 (d), $u^{-n} = 1/u^n$ . We also write $\exp(t)^n$ for $(\exp t)^n$ .

The first proposition collects what the addition rule and the tangent-line bound say at once.

Proposition 8.5.2 (First Consequences of the Addition Rule and the Tangent-Line Bound)

(a) $\exp 0 = 1$ .
(b) $\exp x > 0$ and $\exp(-x) = 1/\exp x$ for every $x \in \mathbb{R}$ .
(c) $\exp$ is strictly increasing on $\mathbb{R}$ .
(d) $\exp x \le 1/(1 - x)$ for every $x \in \mathbb{R}$ with $x < 1$ .
(e) $\exp(nx) = (\exp x)^n$ for every $x \in \mathbb{R}$ and every $n \in \mathbb{Z}$ .

Idea. Part (a) is the addition rule at $x = y = 0$ , with the tangent-line bound to rule out $\exp 0 = 0$ . Part (b) writes $\exp x$ as a square, $\exp(x/2)^2$ . Part (c) factors a difference, as in Example (Method: factor out y − x), but with a product. The addition rule factors $\exp y - \exp x$ as $\exp x \cdot (\exp(y - x) - 1)$ , and the tangent-line bound shows that the bracket is positive. Part (d) is the tangent-line bound at $-x$ , turned upside down.

Proof.

(a) By the addition rule, $\exp 0 = \exp(0 + 0) = (\exp 0)^2$ , so $\exp 0 \cdot (\exp 0 - 1) = 0$ . By the tangent-line bound, $\exp 0 \ge 1 + 0 = 1$ , so $\exp 0 \ne 0$ . A product that is zero with one factor non-zero has its other factor zero, so $\exp 0 - 1 = 0$ .
(b) By the addition rule and (a), $\exp x \cdot \exp(-x) = \exp 0 = 1$ . Hence $\exp x \ne 0$ , and $\exp(-x) = 1/\exp x$ because inverses are unique. The same holds at $x/2$ , so $\exp(x/2) \ne 0$ . By the addition rule, $\exp x = \exp(x/2 + x/2) = (\exp(x/2))^2$ , which is $> 0$ because it is the square of a non-zero number.
(c) Let $x < y$ . By the addition rule, $\exp y = \exp x \cdot \exp(y - x)$ , so $\exp y - \exp x = \exp x \cdot \bigl( \exp(y - x) - 1 \bigr) \ge \exp x \cdot (y - x) > 0 .$ The first inequality is the tangent-line bound at $y - x$ , multiplied by $\exp x > 0$ from (b). The last step holds because a product of two positive numbers is positive, (O4): here $\exp x > 0$ and $y - x > 0$ .
(d) Let $x < 1$ . By the tangent-line bound at $-x$ , $\exp(-x) \ge 1 - x > 0$ . Both numbers are positive, so taking reciprocals reverses the inequality: $1/\exp(-x) \le 1/(1 - x)$ . And $1/\exp(-x) = \exp x$ by (b) applied to $-x$ .
(e) For $n \in \mathbb{N}$ we induct on $n$ . For $n = 0$ , $\exp(0 \cdot x) = \exp 0 = 1 = (\exp x)^0$ by (a). If $\exp(nx) = (\exp x)^n$ , then by the addition rule $\exp((n + 1)x) = \exp(nx + x) = \exp(nx) \exp x = (\exp x)^n \exp x = (\exp x)^{n+1} .$ For $n \in \mathbb{Z}_{>0}$ , (b) and the case just proved give $\exp(-nx) = 1/\exp(nx) = 1/(\exp x)^n = (\exp x)^{-n}$ .

Uses: Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Lemma 4.1.1 (d) (zero products), (a) (inverses are unique). Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (e) (non-zero squares are positive), (c) (multiplying by a positive number). Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.2 (c) (reciprocals reverse order). Proposition 1.5.3 (d) ( $1/u^n = u^{-n}$ ). Logic and Proof, Axiom 4.2.1.

The value property (E3) appears nowhere in this proof. So every part of Proposition 8.5.2 holds for any function that satisfies the addition rule and the tangent-line bound. The main theorem below depends on that.

Quick check. Which of the three borrowed facts does each of the following use: $\exp 0 = 1$ ; $\exp$ is strictly increasing; the inverse of $\exp$ is defined at every point of $(0, \infty)$ ?

Solution. $\exp 0 = 1$ uses the addition rule (E1) and the tangent-line bound (E2), by the proof of (a). Strict increase uses the same two facts, through (b) and (c). The inverse being defined on all of $(0, \infty)$ needs every positive number to be a value, which is the value property (E3). That the inverse is a function at all is (c), so the first two facts enter too.

8.5.4. Squeeze, then close the gap by Archimedes

The main theorem computes $\exp x$ as a supremum. To show $L = \sup S$ , Sets and Functions, Definition 5.3.5 asks that $L$ be an upper bound and that no smaller number be one. The second half often comes from members of $S$ approaching $L$ at a known rate.

A small case first. Let $S$ be the set of the numbers $1 - 1/n$ for $n \in \mathbb{Z}_{>0}$ , and let $L = 1$ . Every member of $S$ is at most $1$ . The member with index $n$ is $L - 1/n$ , so the members come within $1/n$ of $L$ . Given any $L' < 1$ , a large $n$ has $1/n < 1 - L'$ , and then $1 - 1/n > L'$ . So no $L' < 1$ is an upper bound, and $\sup S = 1$ . The method below does the same with any constant $C$ in place of $1$ in $L - 1/n$ .

Example (Method: squeeze, then close the gap by Archimedes)

How it goes. To prove $\sup S = L$ for a set $S \subseteq \mathbb{R}$ :

show that every $s \in S$ satisfies $s \le L$ ;
find a constant $C > 0$ , fixed before $n$ is chosen, and an $N_0 \in \mathbb{Z}_{>0}$ such that for every integer $n \ge N_0$ some $s_n \in S$ satisfies $s_n \ge L - C/n$ ;
given $L' < L$ , put $\varepsilon \mathrel{:=} (L - L')/C > 0$ . By Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.2.3 there is $N$ with $1/n < \varepsilon$ for all $n \ge N$ . For $n \ge \max\{N, N_0\}$ we get $s_n \ge L - C/n > L - C\varepsilon = L'$ .

So no $L' < L$ is an upper bound, and $L$ is the least upper bound.

When to use it. A supremum has been guessed, and the members of $S$ come with an index $n$ and an error estimate of the form $C/n$ .

Common mistakes.

The “constant” $C$ is allowed to depend on $n$ . With $C = n$ , the estimate $s_n \ge L - 1$ says nothing about small gaps.
The constant is $C = 0$ , which leaves $\varepsilon$ undefined. If the estimate seems to come with $C = 0$ , enlarge $C$ : a lower bound $L - C/n$ stays true when $C$ grows.

We know	We want
every $s \in S$ has $s \le L$ ; for $n \ge N_0$ some $s_n \in S$ has $s_n \ge L - C/n$ , with $C > 0$ fixed in advance	no $L' < L$ is an upper bound of $S$
$L' < L$ ; $\varepsilon \mathrel{:=} (L - L')/C > 0$ ; $1/n < \varepsilon$ for $n \ge N$	some $s \in S$ with $s > L'$ : take $s_n$ , $n \ge \max\{N, N_0\}$

The real numbers form a complete ordered field (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 5.5.2). Every complete ordered field is Archimedean (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 6.1.2), so the Archimedean property, which the last step of the method uses, holds in $\mathbb{R}$ .

8.5.5. Two facts determine the exponential

For a fixed $x$ , the numbers $(1 + x/n)^n$ are built from $x$ by arithmetic alone. At $x = 1$ and $n = 2, 3, 4$ they are $9/4 = 2.25$ , $64/27 \approx 2.370$ and $625/256 \approx 2.441$ . The theorem says that these numbers squeeze $\exp x$ from below, and that their supremum is $\exp x$ .

Theorem 8.5.3 (The Exponential Is Determined by Its Addition Rule and Tangent-Line Bound)

Let $x \in \mathbb{R}$ , and let $S_x \mathrel{:=} \Bigl\{ \Bigl(1 + \frac{x}{n}\Bigr)^{n} : n \in \mathbb{Z}_{>0},\ n > \left| x \right| \Bigr\} .$ Then $S_x$ is non-empty and bounded above, and $\exp x = \sup S_x$ . Consequently, if $E \colon \mathbb{R} \to \mathbb{R}$ is any function satisfying the addition rule (E1) and the tangent-line bound (E2) with $E$ in place of $\exp$ , then $E = \exp$ .

Idea (Scratch work). We want two bounds on $(1 + x/n)^n$ : above by $\exp x$ , and below by $\exp x$ minus something of size $C/n$ .

Upper. The tangent-line bound at $x/n$ gives $\exp(x/n) \ge 1 + x/n$ . Raising both sides to the $n$ -th power gives $\exp(x/n)^n \ge (1 + x/n)^n$ , and $\exp(x/n)^n = \exp x$ by Proposition 8.5.2 (e). But raising an inequality to a power is allowed only between non-negative numbers (Lemma 1.5.1 (a)). At $x = -3$ , $n = 2$ , the base $1 + x/n = -1/2$ is negative. So we ask for $n > \left| x \right|$ , which makes $1 + x/n > 0$ .

Lower. Part (d) of the same proposition at $x/n$ gives $\exp(x/n) \le 1/(1 - x/n)$ , valid when $x/n < 1$ , so $\exp x \le (1 - x/n)^{-n}$ . To compare this with $(1 + x/n)^n$ , multiply and divide by $(1 - x/n)^n$ : $\Bigl(1 + \frac{x}{n}\Bigr)^{n} = \Bigl(1 - \frac{x^2}{n^2}\Bigr)^{n} \Bigl(1 - \frac{x}{n}\Bigr)^{-n} .$ The middle factor is close to $1$ : Bernoulli’s inequality gives $(1 - x^2/n^2)^n \ge 1 - x^2/n$ . That is the $C/n$ we need.

We know	We want
the addition rule and the tangent-line bound; Proposition 8.5.2; $n > \left\| x \right\|$	$(1 + x/n)^n \le \exp x$ and $(1 + x/n)^n \ge \exp x - C/n$ , $C > 0$ not depending on $n$
$1 \pm x/n > 0$ ; $\exp x \le (1 - x/n)^{-n}$	$(1 - x^2/n^2)^n \ge 1 - (\text{const})/n$ : Bernoulli

Then Example (Method: squeeze, then close the gap by Archimedes) finishes, with $C \mathrel{:=} (x^2 + 1)\exp x$ . The $+1$ keeps $C > 0$ at $x = 0$ .

Before reading on: why must $n$ exceed $\left| x \right|$ ? Find every place it is used.

Proof. Fix $x \in \mathbb{R}$ . By the Archimedean property there is $N_0 \in \mathbb{Z}_{>0}$ with $N_0 > \left| x \right|$ , so $S_x \ne \varnothing$ . Let $n \in \mathbb{Z}_{>0}$ with $n > \left| x \right|$ . Then $\left| x/n \right| < 1$ , so $1 + x/n > 0$ and $1 - x/n > 0$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.3 (d)).

Upper bound. By the tangent-line bound at $x/n$ , $0 < 1 + x/n \le \exp(x/n)$ . Raising to the $n$ -th power keeps an inequality between non-negative numbers, so $(1 + x/n)^n \le \exp(x/n)^n$ . And $\exp(x/n)^n = \exp(n \cdot x/n) = \exp x$ by Proposition 8.5.2 (e). So $\exp x$ is an upper bound of $S_x$ , and $\sup S_x$ exists.

Lower estimate. Since $x/n < 1$ , Proposition 8.5.2 (d) gives $\exp(x/n) \le 1/(1 - x/n)$ . Both sides are positive, so raising to the $n$ -th power, with part (e) on the left, gives $\exp x = \exp(x/n)^n \le \Bigl(1 - \frac{x}{n}\Bigr)^{-n} .$ Splitting the power of a product, $(1 + x/n)^n = (1 - x^2/n^2)^n (1 - x/n)^{-n}$ . Here $(1 - x^2/n^2)^n > 0$ , since $1 - x^2/n^2 = (1 + x/n)(1 - x/n) > 0$ . Multiply the display by this positive number, and then apply Bernoulli’s inequality (Sets and Functions, Theorem 3.3.7) to $-x^2/n^2 \ge -1$ , using $\exp x > 0$ : $\begin{aligned} \Bigl(1 + \frac{x}{n}\Bigr)^{n} &\ge \Bigl(1 - \frac{x^2}{n^2}\Bigr)^{n} \exp x \ge \Bigl(1 - \frac{x^2}{n}\Bigr) \exp x \\ &\ge \exp x - \frac{(x^2 + 1) \exp x}{n} . \end{aligned}$

Conclusion. Put $C \mathrel{:=} (x^2 + 1)\exp x$ . It depends on $x$ only, and $C > 0$ by Proposition 8.5.2 (b). For every integer $n \ge N_0$ the member $s_n \mathrel{:=} (1 + x/n)^n$ of $S_x$ satisfies $s_n \ge \exp x - C/n$ . By Example (Method: squeeze, then close the gap by Archimedes), $\sup S_x = \exp x$ .

Uniqueness. Let $E$ satisfy the addition rule (E1) and the tangent-line bound (E2), with $E$ in place of $\exp$ . The proof of Proposition 8.5.2 used only these two facts and field and order rules, so its conclusions hold for $E$ . The argument above used only those conclusions, the tangent-line bound, and facts about $\mathbb{R}$ . So $E(x) = \sup S_x = \exp x$ for every $x \in \mathbb{R}$ . Two functions with the same domain, codomain and values are equal, so $E = \exp$ .

Uses: Lemma 1.5.1 (a) (powers keep order). Sets and Functions, Proposition 3.3.4 (c) (power of a product). Sets and Functions, Proposition 2.1.1 (equal functions).

The set $S_x$ never mentions $\exp$ , so the theorem is a formula: $\exp x$ is pinned between computable numbers, as the next definition shows at $x = 1$ . And the three borrowed facts describe a function that the first two already determine.

8.5.6. The number e

Definition 8.5.1 (Euler’s Number)

$e \mathrel{:=} \exp 1$ .

By the proof of Theorem 8.5.3 at $x = 1$ , for every integer $n \ge 2$ , $\Bigl(1 + \frac1n\Bigr)^{n} \le e \le \Bigl(1 - \frac1n\Bigr)^{-n} .$ At $n = 2$ this reads $9/4 \le e \le 4$ , so $e > 2$ . Exercise B1 below squeezes harder.

Now we can say more about what fails without the tangent-line bound, beyond the constant function $1$ . The function $F(x) \mathrel{:=} \exp(2x)$ satisfies the addition rule, since $F(x + y) = \exp(2x)\exp(2y)$ . It also has the value property, since $\exp t = y$ gives $F(t/2) = y$ . It is not $\exp$ : $F(1) = \exp 2 = (\exp 1)^2 = e \cdot e$ by Proposition 8.5.2 (e), and $e \cdot e \ne e$ because $e > 1$ . So, by Theorem 8.5.3, it must fail the tangent-line bound. It does, at $x = -1/2$ : $F(-1/2) = \exp(-1) = 1/e < 1/2 = 1 + (-1/2)$ , as $e > 2$ . Without the tangent-line bound, the addition rule and the value property leave the speed of the function free.

8.5.7. The logarithm

By Proposition 8.5.2 (b) and (c), $\exp$ is a strictly increasing map from $\mathbb{R}$ into $(0, \infty)$ . By the value property, every point of $(0, \infty)$ is a value. So $\exp \colon \mathbb{R} \to (0, \infty)$ is a bijection.

The logarithm undoes the exponential.

Definition 8.5.2 (The Logarithm)

The logarithm $\log \colon (0, \infty) \to \mathbb{R}$ is the inverse of the bijection $\exp \colon \mathbb{R} \to (0, \infty)$ : for $y > 0$ , $\log y$ is the unique $x \in \mathbb{R}$ with $\exp x = y$ .

In words: the existence of $x$ is the value property. Its uniqueness holds because a strictly increasing map is injective (Sets and Functions, Proposition 2.2.3). So $\exp(\log y) = y$ for every $y > 0$ and $\log(\exp x) = x$ for every $x \in \mathbb{R}$ . By Theorem 8.2.3, $\log$ is strictly increasing, and its graph is the graph of $\exp$ with the coordinates swapped, that is, reflected in the line $y = x$ . We write $\log$ and never $\ln$ .

Diagram. Labels: x, y, y = x, y = 1 + x, 1

The picture shows $\exp$ above the line $y = 1 + x$ , touching it at $(0, 1)$ , and $\log$ as its mirror image in $y = x$ , through $(1, 0)$ . It illustrates; no proof below reads anything off it.

Quick check. What are $\log(\exp 2)$ and $\log(1/e)$ ?

Solution. $\log(\exp 2) = 2$ , since $\log$ undoes $\exp$ . And $\exp(-1) = 1/\exp 1 = 1/e$ by Proposition 8.5.2 (b), so $\log(1/e) = -1$ .

Each rule for $\log$ comes from a rule for $\exp$ : apply $\exp$ to both sides and use that $\exp$ is injective.

Proposition 8.5.4 (Rules for the Logarithm)

Let $x, y \in \mathbb{R}$ with $x, y > 0$ .

(a) $\log 1 = 0$ and $\log e = 1$ .
(b) $\log(xy) = \log x + \log y$ and $\log(1/x) = -\log x$ .
(c) $1 - 1/x \le \log x \le x - 1$ , and each inequality is an equality only at $x = 1$ .

Idea. For (a) and (b), compute $\exp$ of each side. For (c), the tangent-line bound at $t = \log x$ gives $x \ge 1 + \log x$ , and at $t = \log(1/x)$ it gives the other half. Strictness needs $\exp t > 1 + t$ for $t \ne 0$ , which the three facts do not state. We get it by writing $\exp t$ as a square, $\exp(t/2)^2 \ge (1 + t/2)^2 = 1 + t + t^2/4$ , valid while $1 + t/2 \ge 0$ .

Proof.

(a) We have $\exp 0 = 1$ by Proposition 8.5.2 (a), and $\exp 1 = e$ by the definition of $e$ . So $\log 1 = 0$ and $\log e = 1$ , by the definition of $\log$ .
(b) By the addition rule, $\exp(\log x + \log y) = \exp(\log x)\exp(\log y) = xy$ , so $\log(xy) = \log x + \log y$ by the definition of $\log$ . Then $\log x + \log(1/x) = \log 1 = 0$ by (a).
(c) First, $\exp t > 1 + t$ for every $t \ne 0$ . Case $t \ge -2$ . Then $1 + t/2 \ge 0$ , and $\exp(t/2) \ge 1 + t/2$ by the tangent-line bound. Squaring keeps an inequality between non-negative numbers, so by the addition rule $\exp t = \exp(t/2)^2 \ge \Bigl(1 + \frac t2\Bigr)^{2} = 1 + t + \frac{t^2}{4} > 1 + t ,$ the last step because $t^2 > 0$ . Case $t < -2$ . Then $1 + t < -1 < 0 < \exp t$ by Proposition 8.5.2 (b).

Now the tangent-line bound at $t = \log x$ gives $x = \exp(\log x) \ge 1 + \log x$ , that is, $\log x \le x - 1$ . By the strict inequality just proved, it is strict when $\log x \ne 0$ , that is, when $x \ne 1$ . By (b), $\log(1/x) = -\log x$ . The tangent-line bound at $t = -\log x$ gives $1/x \ge 1 - \log x$ , that is, $\log x \ge 1 - 1/x$ , again strict when $x \ne 1$ . At $x = 1$ all three numbers are $0$ .

Uses: Definition 8.5.1 ( $e = \exp 1$ ). Definition 8.5.2 ( $\log$ inverts $\exp$ ). Lemma 1.5.1 (a) (squares keep order). Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (e) (non-zero squares are positive).

Part (c) gives numbers: $1/2 < \log 2 < 1$ , for example.

8.5.8. Real powers

A real power is the exponential of a multiple of a logarithm.

For example, the rule below sets $2^{\sqrt 2} \mathrel{:=} \exp(\sqrt 2 \log 2)$ and $8^{1/3} \mathrel{:=} \exp(\tfrac13 \log 8)$ . The second should be $2$ , the cube root of $8$ , and we will check that it is.

Definition 8.5.3 (Real Powers)

Let $a, x \in \mathbb{R}$ .

If $a > 0$ , then $a^x \mathrel{:=} \exp(x \log a)$ .
If $a = 0$ and $x > 0$ , then $0^x \mathrel{:=} 0$ .

No other pair $(a, x)$ is given a value by this definition; in particular there are no real powers of a negative base.

In words: to raise $a$ to the power $x$ , take the logarithm, multiply by $x$ , and exponentiate. By Definition 8.5.2, $\log(a^x) = x \log a$ for $a > 0$ . This is the definition with $\log$ applied to both sides, not a theorem.

The symbol $a^x$ was already in use for $x \in \mathbb{N}$ (Sets and Functions, Definition 3.3.2), and Chapter 1 gave $\sqrt[n]{a}$ and $a^{m/n}$ . The next proposition shows that the new definition agrees with both wherever they overlap, so no symbol now has two meanings.

Second readingBase zero and negative bases

Base zero. For $x = n \in \mathbb{Z}_{>0}$ both definitions give $0^n = 0$ . For $x = 0$ , the definition of real powers says nothing. The symbol $0^0$ keeps the value $1$ that Sets and Functions, Definition 3.3.2 gives it as a natural-number power. For $x < 0$ , $0^x$ would have to be $1/0^{-x} = 1/0$ , so it is left undefined.

Negative bases. $\log a$ is defined only for $a > 0$ , so the formula $\exp(x \log a)$ cannot start. One could set $(-8)^{1/3} \mathrel{:=} -2$ , but the rule $(a^x)^y = a^{xy}$ would then break. With $x = 2$ and $y = 1/6$ it would give $(-8)^{1/3} = ((-8)^2)^{1/6} = 64^{1/6} = 2$ , not $-2$ . The real cube root of a negative number is a separate object, $\sqrt[3]{-8} = -2$ of Example (Roots as inverse functions), and it keeps its own symbol.

Quick check. Does Definition 8.5.3 give $0^0$ a value? Is $(1/2)^{-1} = 2$ ?

Solution. No: base $0$ is covered only for $x > 0$ . The symbol $0^0$ keeps its natural-number value $1$ from Sets and Functions, Definition 3.3.2, which this definition does not touch. Yes: $(1/2)^{-1} = \exp(-\log(1/2)) = \exp(\log 2) = 2$ , since $-\log(1/2) = \log 2$ by Proposition 8.5.4 (b).

Proposition 8.5.5 (Real Powers Agree with Roots and Integer Powers)

Let $a \in \mathbb{R}$ with $a > 0$ .

(a) For $m \in \mathbb{Z}$ and $n \in \mathbb{Z}_{>0}$ , the real power $a^{m/n}$ is $(\sqrt[n]{a})^m$ , the rational power of Definition 1.5.2, where $(\sqrt[n]{a})^m$ is the integer power of Chapter 1.
(b) In particular, for $n = 1$ , the real power $a^m$ is the integer power $a^m$ (for $m \ge 0$ , the power of Sets and Functions, Definition 3.3.2).
(c) $e^x = \exp x$ for every $x \in \mathbb{R}$ .

Idea. Show that $a^{1/n}$ is a non-negative number whose $n$ -th power is $a$ ; the uniqueness of roots makes it $\sqrt[n]{a}$ . Then take the $m$ -th power.

Proof.

(a) By the definition of real powers, $c \mathrel{:=} a^{1/n} = \exp((1/n) \log a)$ , and $c > 0$ by Proposition 8.5.2 (b). By part (e) of the same proposition, with the integer $n$ , $c^n = \exp\Bigl(n \cdot \frac 1n \log a\Bigr) = \exp(\log a) = a .$ So $c \ge 0$ and $c^n = a$ . By the uniqueness in Lemma 1.5.1 (b), $c = \sqrt[n]{a}$ . By part (e) again, now with $(1/n) \log a$ in place of $x$ and the integer $m$ , $a^{m/n} = \exp\Bigl(m \cdot \frac 1n \log a\Bigr) = c^m = (\sqrt[n]{a})^m ,$ and part (e) states $c^m$ with the integer powers of Definition 1.5.1 when $m < 0$ .
(b) At $n = 1$ , $\sqrt[1]{a}$ is the unique $b \ge 0$ with $b^1 = a$ , which is $a$ itself. So (a) gives $a^{m/1} = a^m$ .
(c) By Proposition 8.5.4 (a), $e^x = \exp(x \log e) = \exp(x \cdot 1) = \exp x$ .

Uses: Definition 8.5.3 ( $a^x = \exp(x \log a)$ ).

From now on $e^x$ and $\exp x$ are the same thing, and we use whichever reads better.

Quick check. Is $4^{1/2} = 2$ ? Answer from Proposition 8.5.5.

Solution. Yes. By (a) with $m = 1$ , $n = 2$ , $4^{1/2} = (\sqrt 4)^1 = \sqrt 4$ , the unique non-negative real number whose square is $4$ . Since $2 \ge 0$ and $2^2 = 4$ , that number is $2$ .

The laws of exponents now come in one line each, because each is a law of $\exp$ or of $\log$ in disguise. For whole-number exponents they are school rules: $2^3 \cdot 2^2 = 8 \cdot 4 = 32 = 2^5$ , and $(2^3)^2 = 64 = 2^6$ . The proposition says that they hold for every real exponent.

Proposition 8.5.6 (Laws of Real Powers)

Let $a, b, x, y \in \mathbb{R}$ with $a, b > 0$ .

(a) $a^{x + y} = a^x a^y$ and $a^{-x} = 1/a^x$ .
(b) $(a^x)^y = a^{xy}$ .
(c) $(ab)^x = a^x b^x$ .
(d) $a^x > 0$ .
(e) As a function of $x$ on $\mathbb{R}$ : $x \mapsto a^x$ is strictly increasing if $a > 1$ , strictly decreasing if $a < 1$ , and constant $1$ if $a = 1$ .
(f) As a function of $a$ on $(0, \infty)$ : $a \mapsto a^x$ is strictly increasing if $x > 0$ and strictly decreasing if $x < 0$ .

Idea. Write each power as $\exp$ of a product and use the addition rule, the log rules, and the monotonicity of $\exp$ and $\log$ .

Proof.

(a) $a^{x+y} = \exp(x \log a + y \log a) = a^x a^y$ by the addition rule. Also $a^{-x} = \exp(-x \log a) = 1/a^x$ by Proposition 8.5.2 (b).
(b) $a^x > 0$ by (d), and $\log(a^x) = x \log a$ , so $(a^x)^y = \exp(y x \log a) = a^{xy}$ .
(c) $(ab)^x = \exp(x \log a + x \log b) = a^x b^x$ , by Proposition 8.5.4 (b) and the addition rule.
(d) This is Proposition 8.5.2 (b).
(e) If $a > 1$ , then $\log a > \log 1 = 0$ , since $\log$ is strictly increasing. So $x \mapsto x \log a$ is strictly increasing, because multiplying by a positive number keeps order. Composing with the strictly increasing $\exp$ gives a strictly increasing function (Proposition 8.2.1 (e)). If $a < 1$ , then $\log a < 0$ , and $x \mapsto x \log a$ is strictly decreasing, because multiplying by a negative number reverses order. The composite is strictly decreasing by the same rule. If $a = 1$ , $a^x = \exp 0 = 1$ .
(f) Let $0 < a < b$ . Then $\log a < \log b$ . If $x > 0$ , then $x \log a < x \log b$ , and $\exp$ keeps the order. If $x < 0$ , the order is reversed and $\exp$ keeps the reversed order.

Uses: Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (c), (d) (multiplying by a sign). Definition 8.5.2 ( $\log$ increasing). Proposition 8.5.4 (a) ( $\log 1 = 0$ ).

Quick check. Is $2^{\sqrt 2} > 2$ ? Is $(\sqrt 2)^{\sqrt 2} < 2$ ?

Solution. Yes to both. Since $1^2 = 1 < 2 = (\sqrt 2)^2$ , we have $\sqrt 2 > 1$ , because squares of non-negative numbers keep order. The map $x \mapsto 2^x$ is strictly increasing by Proposition 8.5.6 (e), so $2^{\sqrt 2} > 2^1 = 2$ . For the second, $\sqrt 2 = 2^{1/2}$ by Proposition 8.5.5 (a). So $(\sqrt 2)^{\sqrt 2} = 2^{\sqrt 2/2}$ , by Proposition 8.5.6 (b). Since $(\sqrt 2)^2 = 2 < 4 = 2^2$ , $\sqrt 2 < 2$ , so $\sqrt 2/2 < 1$ and $2^{\sqrt 2/2} < 2^1 = 2$ .

Uses: Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (g) (comparing squares).

8.5.9. The exponential outruns every power

How fast does $\exp x$ grow? The tangent-line bound gives only $\exp x \ge 1 + x$ , which a linear function matches. Raising the tangent-line bound to a power, as in the proof of Theorem 8.5.3, does much better.

A small case first. Split $10$ into two halves. By the addition rule, $\exp 10 = (\exp 5)^2$ , and the tangent-line bound gives $\exp 5 \ge 6$ , so $\exp 10 \ge 36$ . The tangent-line bound at $10$ alone gives only $\exp 10 \ge 11$ . Splitting into more pieces does better still.

Proposition 8.5.7 (The Exponential Beats Every Power)

(a) For every $x > 0$ and every $k \in \mathbb{N}$ , $\exp x > \bigl(x/(k + 1)\bigr)^{k+1}$ . Hence, for every $M > 0$ , $\exp x > M x^k$ for all $x > M (k + 1)^{k+1}$ .
(b) For every $x > 0$ and every $m \in \mathbb{Z}_{>0}$ , $\log x < m x^{1/m}$ . Hence, for every $\varepsilon > 0$ , $\log x < \varepsilon x$ for all $x > 4/\varepsilon^2$ .

Idea. For (a), split $x$ into $k + 1$ equal parts and apply the tangent-line bound to each: $\exp x = \exp(x/(k + 1))^{k+1} \ge (1 + x/(k + 1))^{k+1}$ . Dropping the $1$ loses a little, and gives a strict inequality. For (b), apply $\log t \le t - 1$ not at $x$ but at $x^{1/m}$ , which is much smaller.

Proof.

(a) Let $x > 0$ , $k \in \mathbb{N}$ , $t \mathrel{:=} x/(k + 1) > 0$ . By the tangent-line bound, $\exp t \ge 1 + t > t \ge 0$ . Raising to the power $k + 1$ keeps a strict inequality between non-negative numbers, so by Proposition 8.5.2 (e), $\exp x = (\exp t)^{k+1} > t^{k+1} = \frac{x^{k+1}}{(k + 1)^{k+1}} = x^k \cdot \frac{x}{(k + 1)^{k+1}} .$ If $M > 0$ and $x > M(k + 1)^{k+1}$ , then $x > 0$ , $x^k > 0$ and $x/(k + 1)^{k+1} > M$ , so the right side exceeds $M x^k$ .
(b) Let $x > 0$ and $m \in \mathbb{Z}_{>0}$ . By the definition of real powers, $\log(x^{1/m}) = (1/m) \log x$ . So, by Proposition 8.5.4 (c) at $x^{1/m} > 0$ , $\log x = m \log(x^{1/m}) \le m (x^{1/m} - 1) < m x^{1/m} .$ Now let $\varepsilon > 0$ and $x > 4/\varepsilon^2$ , and take $m = 2$ . Here $x^{1/2} = \sqrt x$ by Proposition 8.5.5 (a). Since $\sqrt x \ge 0$ and $(\sqrt x)^2 = x > (2/\varepsilon)^2$ , we get $\sqrt x > 2/\varepsilon$ , because squares of non-negative numbers keep order. So $2 < \varepsilon \sqrt x$ and $2\sqrt x < \varepsilon \sqrt x \sqrt x = \varepsilon x$ . Hence $\log x < 2\sqrt x < \varepsilon x$ .

Uses: Lemma 1.5.1 (a) (powers keep order). Sets and Functions, Proposition 3.3.4 (c) (power of a product). Definition 1.5.1, Proposition 1.5.3 (d) ( $(1/u)^n = 1/u^n$ ). Definition 8.5.3 ( $\log(a^x) = x \log a$ ). Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (g) (comparing squares).

The thresholds are explicit: $\exp x > 1000\, x^2$ once $x > 1000 \cdot 27 = 27{,}000$ , and $\log x < x/100$ once $x > 40{,}000$ . They are far from the best, and need not be good. They need to exist, depending on $M$ , $k$ or $\varepsilon$ only, never on $x$ .

8.5.10. Exercises

A. Check your understanding

Exercise (A1)

These questions are about the proof of Theorem 8.5.3.

(a) Which of the addition rule (E1), the tangent-line bound (E2) and the value property (E3) does the proof use, including through Proposition 8.5.2?
(b) Where does Bernoulli’s inequality enter, and at which real number is it applied?
(c) Write out the two bounds on $e$ that the proof gives at $x = 1$ , $n = 2$ .
(d) Take $x = -3$ and $n = 2$ . Which step of the upper-bound argument fails? Is its conclusion $(1 + x/n)^n \le \exp x$ still true for this pair?

Solution.

(a) Only the addition rule (E1) and the tangent-line bound (E2). The proof of Proposition 8.5.2 uses these two and field and order rules. The proof of the theorem adds the tangent-line bound at $x/n$ . The value property (E3) is not used, which is why the uniqueness sentence asks only for the first two.
(b) In the lower estimate, to bound $(1 - x^2/n^2)^n$ below by $1 - x^2/n$ . It is applied at $-x^2/n^2$ , which is $\ge -1$ because $n > \left| x \right|$ .
(c) $(1 + 1/2)^2 = 9/4 \le e \le (1 - 1/2)^{-2} = 4$ .
(d) Here $n = 2 < 3 = \left| x \right|$ , and $1 + x/n = -1/2 < 0$ . The step “raise $1 + x/n \le \exp(x/n)$ to the $n$ -th power” fails: Lemma 1.5.1 (a) needs a non-negative smaller side. The conclusion fails too: $(-1/2)^2 = 1/4$ , while $\exp(-3) = 1/\exp 3 < 1/4$ , because $\exp 3 > 1 + 3 = 4$ . The inequality $\exp 3 > 4$ is strict by the first step of the proof of Proposition 8.5.4 (c).

Exercise (A2)

True or false? Give a reason or a counterexample.

(a) $\exp(2x) = (\exp x)^2$ for every $x \in \mathbb{R}$ .
(b) $\log(x + y) = \log x + \log y$ for all $x, y > 0$ .
(c) Definition 8.5.3 defines $(-8)^{1/3}$ .
(d) $a^x b^y = (ab)^{x + y}$ for all $a, b > 0$ and $x, y \in \mathbb{R}$ .

Solution.

(a) True, by the addition rule (E1) with $y = x$ , or by Proposition 8.5.2 (e) with $n = 2$ .
(b) False. At $x = y = 1$ the left side is $\log 2$ and the right side is $0$ . But $\log 2 > \log 1 = 0$ , since $\log$ is strictly increasing.
(c) False: the base $-8$ is negative, and Definition 8.5.3 gives no real powers of a negative base. The real number $\sqrt[3]{-8} = -2$ is the odd root of Example (Roots as inverse functions), a different symbol.
(d) False. With $a = 1$ , $b = 2$ , $x = 1$ , $y = 0$ : the left side is $1^1 \cdot 2^0 = 1$ , the right side $2^1 = 2$ (values by Proposition 8.5.5 (b)).

B. Practice

Exercise (B1)

Prove that $2.5 < e < 3$ .

Hint: one value of $n$ in the bounds after Definition 8.5.1 does both.

Solution. Take $n = 6$ . The lower bound gives $e \ge \Bigl(\frac76\Bigr)^{6} = \frac{117649}{46656} > \frac52 ,$ because $2 \cdot 117649 = 235298 > 233280 = 5 \cdot 46656$ . The upper bound gives $e \le \Bigl(1 - \frac16\Bigr)^{-6} = \Bigl(\frac65\Bigr)^{6} = \frac{46656}{15625} < 3 ,$ because $46656 < 46875 = 3 \cdot 15625$ . Hence $2.5 < e < 3$ . Smaller $n$ is not enough: at $n = 2, 3, 4, 5$ the upper bound $(n/(n - 1))^n$ is $4$ , $27/8$ , $256/81$ , $3125/1024$ , all above $3$ .

Exercise (B2: Prove or disprove)

Prove each true statement and give a counterexample to each false one.

(a) $\exp x \ge 1 + x + x^2/2$ for every $x \ge 0$ .
(b) $\log x \le x/e$ for every $x > 0$ .
(c) If $0 < a \le b$ , then $a^x \le b^x$ for every $x \in \mathbb{R}$ .
(d) There is a polynomial function $p$ on $\mathbb{R}$ (Definition 8.1.3) with $p(x) = \exp x$ for every $x \in \mathbb{R}$ .

Solution.

(a) True. Let $x \ge 0$ and $n \in \mathbb{Z}_{>0}$ with $n > x$ and $n \ge 2$ . By the binomial theorem in the commutative ring $\mathbb{R}$ (Theorem 1.2.3), $(1 + x/n)^n$ is a sum of non-negative terms. Among them are $1$ , $n \cdot x/n = x$ and $\binom n2 x^2/n^2 = (1 - 1/n) x^2/2$ . Dropping the others, $\exp x \ge \Bigl(1 + \frac xn\Bigr)^{n} \ge 1 + x + \frac{x^2}{2} - \frac{x^2}{2n} ,$ where the first inequality is the upper-bound half of Theorem 8.5.3. Suppose $\delta \mathrel{:=} 1 + x + x^2/2 - \exp x > 0$ . Then $x \ne 0$ , since at $x = 0$ we have $\delta = 0$ . The display gives $x^2/(2n) \ge \delta$ for every integer $n \ge 2$ with $n > x$ . By Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.2.3 there is $n \ge 2$ with $n > x$ and $1/n < 2\delta/x^2$ , a contradiction. So $\delta \le 0$ .
(b) True. By Proposition 8.5.4 (b), (a) and (c), $\log x = \log(e \cdot x/e) = 1 + \log(x/e) \le 1 + (x/e - 1) = x/e$ .
(c) False. Take $a = 1$ , $b = 2$ , $x = -1$ : $1^{-1} = 1 > 1/2 = 2^{-1}$ . The statement holds for $x \ge 0$ (Proposition 8.5.6 (f), with equality at $x = 0$ ) and reverses for $x < 0$ .
(d) False. Suppose $p(x) = \sum_{i=0}^{d} a_i x^i = \exp x$ for all $x$ . Put $S \mathrel{:=} \sum_{i<d} \left| a_i \right|$ and $M \mathrel{:=} S + \left| a_d \right| + 1 \ge 1$ . For $x \ge 1$ we have $x^i \le x^d$ for $i \le d$ . So the triangle inequality gives $\left| p(x) \right| \le (S + \left| a_d \right|) x^d < M x^d$ . By Proposition 8.5.7 (a) with $k = d$ , any $x > M(d + 1)^{d+1}$ , which is $\ge 1$ , has $\exp x > M x^d > \left| p(x) \right| \ge p(x)$ , a contradiction.

Uses: Sets and Functions, Proposition 3.3.4 (e) (powers of $x \ge 1$ ). Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.3 (e) (triangle inequality).

Exercise (B3: Put the proof in order)

Here are lines of the lower estimate in the proof of Theorem 8.5.3, shuffled, with two lines that do not belong. Write an order of the correct lines that makes a proof, and say why each distractor fails.

(i) Fix $x \in \mathbb{R}$ and an integer $n > \left| x \right|$ .
(ii) Then $1 + x/n > 0$ and $1 - x/n > 0$ .
(iii) By Proposition 8.5.2 (d), $\exp(x/n) \le 1/(1 - x/n)$ , so $\exp x \le (1 - x/n)^{-n}$ .
(iv) $(1 + x/n)^n = (1 - x^2/n^2)^n (1 - x/n)^{-n}$ .
(v) So $(1 + x/n)^n \ge (1 - x^2/n^2)^n \exp x$ .
(vi) Bernoulli’s inequality at $-x^2/n^2 \ge -1$ gives $(1 - x^2/n^2)^n \ge 1 - x^2/n$ .
(vii) Hence $(1 + x/n)^n \ge \exp x - x^2 \exp x / n$ .
(viii) (D1) By the tangent-line bound, $\exp x \ge 1 + x$ , so $(1 + x/n)^n \ge 1 + x$ .
(ix) (D2) Bernoulli’s inequality at $x^2/n^2$ gives $(1 - x^2/n^2)^n \ge 1 + x^2/n$ .

Solution. Each line may come only after the lines it uses. Line (ii) uses (i). Line (iii) uses (ii), since it needs $x/n < 1$ and positive sides. Line (iv) uses (ii), since it divides by $1 - x/n \ne 0$ . Line (v) uses (iii) and (iv). Line (vi) uses only (i), since it needs $x^2/n^2 \le 1$ . Line (vii) uses (v) and (vi). Every order respecting these constraints is accepted. One is (i), (ii), (iii), (iv), (v), (vi), (vii); another is (i), (vi), (ii), (iv), (iii), (v), (vii).

(D1) bounds the wrong thing: the goal is a lower bound by $\exp x - C/n$ , and $1 + x$ is not one. Its “so” is also unjustified: $(1 + x/n)^n \ge 1 + x$ is Bernoulli’s inequality, not a consequence of $\exp x \ge 1 + x$ . (D2) has the wrong sign: Bernoulli’s inequality at $u = -x^2/n^2$ gives $1 + nu = 1 - x^2/n$ . The claimed bound is false, since for $x \ne 0$ the left side is $< 1$ and the right side $> 1$ .

C. Going deeper

Exercise (C1: Find the error)

Here is a proof that $\exp$ is strictly increasing: “Let $x < y$ . By the addition rule, $\exp y - \exp x = \exp(y - x)$ , and by the tangent-line bound, $\exp(y - x) \ge 1 + (y - x) > 0$ . So $\exp y > \exp x$ .” Is the conclusion true? Does the argument prove it? Name the line that fails, and show that it fails with numbers.

Solution. The conclusion is true: it is Proposition 8.5.2 (c). The argument does not prove it. The first equality is false: the addition rule (E1) gives the quotient $\exp y / \exp x = \exp(y - x)$ , not the difference. At $x = 0$ , $y = 1$ : $\exp 1 - \exp 0 = e - 1$ , while $\exp(1 - 0) = e$ , and $e - 1 \ne e$ . The correct line is $\exp y - \exp x = \exp x \cdot (\exp(y - x) - 1)$ . After it, the tangent-line bound gives $\exp(y - x) - 1 \ge y - x > 0$ , and multiplying by $\exp x > 0$ finishes.

Exercise (C2: Chapter check: the logarithm is slower than every root)

(a) Prove that for every $\varepsilon > 0$ there is $X$ such that $\log x < \varepsilon \sqrt x$ for every $x > X$ .
(b) Let $m \in \mathbb{Z}_{>0}$ . Prove that for every $\varepsilon > 0$ there is $X$ such that $\log x < \varepsilon x^{1/m}$ for every $x > X$ .

Solution (Hint 1). For (a), compare $\log x$ with a root of $x$ smaller than $\sqrt x$ , using Proposition 8.5.7 (b).

Solution (Hint 2). For (b), put $y \mathrel{:=} x^{2/m}$ . What is $\log x$ in terms of $\log y$ ?

Solution.

(a) Let $\varepsilon > 0$ and $X \mathrel{:=} (4/\varepsilon)^4 = 256/\varepsilon^4$ ; it depends on $\varepsilon$ only. Let $x > X$ , and put $s \mathrel{:=} x^{1/4}$ . By Proposition 8.5.7 (b) with $m = 4$ , $\log x < 4s$ . By Proposition 8.5.5 (a), $s = \sqrt[4]{x}$ , so $s \ge 0$ and $s^4 = x > (4/\varepsilon)^4$ . Hence $s > 4/\varepsilon$ , because fourth powers of non-negative numbers keep order. So $4 < \varepsilon s$ and $4s < \varepsilon s \cdot s$ . Finally $s \cdot s = x^{1/4 + 1/4} = x^{1/2} = \sqrt x$ , by the addition law for real powers and Proposition 8.5.5 (a). Hence $\log x < 4s < \varepsilon \sqrt x$ .
(b) Let $\varepsilon > 0$ and put $\delta \mathrel{:=} 2\varepsilon/m$ . Let $Y \mathrel{:=} 256/\delta^4 = 16m^4/\varepsilon^4$ , the threshold of (a) for $\delta$ , and $X \mathrel{:=} Y^{m/2}$ ; both depend on $\varepsilon$ and $m$ only. Let $x > X$ and $y \mathrel{:=} x^{2/m} > 0$ . The map $a \mapsto a^{2/m}$ is strictly increasing on $(0, \infty)$ by Proposition 8.5.6 (f). So $y > X^{2/m} = Y^{(m/2)(2/m)} = Y^1 = Y$ , by part (b) of the same proposition and Proposition 8.5.5 (b). By (a) applied to $y$ and $\delta$ , $\log y < \delta \sqrt y$ . Here $\sqrt y = y^{1/2} = x^{(2/m)(1/2)} = x^{1/m}$ , by Proposition 8.5.5 (a) and the law $(a^x)^y = a^{xy}$ . By the definition of real powers, $\log y = (2/m) \log x$ , so $\log x = \frac m2 \log y < \frac m2 \, \delta \, x^{1/m} = \varepsilon x^{1/m} .$

What a complete answer needs (10 points)

3: a bound $\log x \le c\, x^{1/k}$ with $k > 2$ , with the result or log rule it comes from named.
3: an explicit $X$ depending on $\varepsilon$ only, and a check that $x > X$ gives $\log x < \varepsilon \sqrt x$ .
2: passes from $x > X$ to the bound in (b), for example by the substitution $y = x^{2/m}$ with the monotonicity of $a \mapsto a^{2/m}$ cited, or directly by Proposition 8.5.7 (b) with $2m$ and $X = (2m/\varepsilon)^{2m}$ .
2: the quantifiers in the order $\varepsilon$ , then $X$ , then every $x > X$ .

A route through Proposition 8.5.7 (a) (put $x = \exp t$ ) earns the same points step for step.

Slips that lose the step (0 for the step): $X$ chosen depending on $x$ ; any use of derivatives, l’Hôpital’s rule or limits of functions.

Uses: Lemma 1.5.1 (a) (powers keep order). Proposition 8.5.6 (a), (b) ( $a^{x+y} = a^x a^y$ , $(a^x)^y = a^{xy}$ ). Definition 8.5.3 ( $\log(a^x) = x \log a$ ).