The logarithm and the exponential

To prove existence, we define a candidate and show it satisfies all the properties. Let

To prove property iv, we change variables $u=yt$ to obtain

Let us prove iii. Property ii together with the fact that $L'(x) = \nicefrac{1}{x} > 0$ for $x > 0\text{,}$ implies that $L$ is strictly increasing and hence one-to-one. Let us show $L$ is onto. As $\nicefrac{1}{t} \geq \nicefrac{1}{2}$ when $t \in [1,2]\text{,}$

By induction, iv implies that for $n \in \N\text{,}$

Given $y > 0\text{,}$ by the Archimedean property of the real numbers (notice $L(2) > 0$), there is an $n \in \N$ such that $L(2^n) > y\text{.}$ The intermediate value theorem gives an $x_1 \in (1,2^n)$ such that $L(x_1) = y\text{.}$ Thus $(0,\infty)$ is in the image of $L\text{.}$ As $L$ is increasing, $L(x) > y$ for all $x > 2^n\text{,}$ and so

Next $0 = L(\nicefrac{x}{x}) = L(x) + L(\nicefrac{1}{x})\text{,}$ and so $L(x) = - L(\nicefrac{1}{x})\text{.}$ Using $x=2^{-n}\text{,}$ we obtain as above that $L$ achieves all negative numbers. And

In the limits, note that only $x > 0$ are in the domain of $L\text{.}$

Let us prove v. Fix $x > 0\text{.}$ As above, iv implies $L(x^n) = n L(x)$ for all $n \in \N\text{.}$ We already found that $L(x) = - L(\nicefrac{1}{x})\text{,}$ so $L(x^{-n}) = - L(x^n) = -n L(x)\text{.}$ Then for $m \in \N$

Putting everything together for $n \in \Z$ and $m \in \N\text{,}$ we have $L(x^{n/m}) = n L(x^{1/m}) = (\nicefrac{n}{m}) L(x)\text{.}$

Uniqueness follows using properties i and ii. Via the first form of the fundamental theorem of calculus (Theorem 5.3.1),

is the unique function such that $L(1) = 0$ and $L'(x) = \nicefrac{1}{x}\text{.}$

Property ii follows via the inverse function theorem, in particular via Lemma 4.4.1: $L$ satisfies all the hypotheses of the lemma, and hence

Let us look at property iii. The function $E$ is strictly increasing since $E'(x) = E(x) > 0\text{.}$ As $E$ is the inverse of $L\text{,}$ it must also be bijective. To find the limits, we use that $E$ is strictly increasing and onto $(0,\infty)\text{.}$ For every $M > 0\text{,}$ there is an $x_0$ such that $E(x_0) = M$ and $E(x) \geq M$ for all $x \geq x_0\text{.}$ Similarly, for every $\epsilon > 0\text{,}$ there is an $x_0$ such that $E(x_0) = \epsilon$ and $E(x) < \epsilon$ for all $x < x_0\text{.}$ Therefore,

To prove property iv, we use the corresponding property for the logarithm. Take $x, y \in \R\text{.}$ As $L$ is bijective, find $a$ and $b$ such that $x = L(a)$ and $y = L(b)\text{.}$ Then

Property v also follows from the corresponding property of $L\text{.}$ Given $x \in \R\text{,}$ let $a$ be such that $x = L(a)$ and

Uniqueness follows from i and ii. Let $E$ and $F$ be two functions satisfying i and ii.

Therefore, by Proposition 4.2.6, $F(x)E(-x) = F(0)E(-0) = 1$ for all $x \in \R\text{.}$ Doing the computation with $F = E\text{,}$ we obtain $E(x)E(-x) = 1\text{.}$ Then

Finally, $E(-x) \neq 0$

¹

$E$ is a function into $(0,\infty)$ after all. However, $E(-x) \neq 0$ also follows from $E(x)E(-x) = 1\text{.}$ Therefore, we can prove uniqueness of $E$ given i and ii, even for functions $E \colon \R \to \R\text{.}$

for all $x \in \R\text{.}$ So $F(x)-E(x) = 0$ for all $x\text{,}$ and we are done.