Equivalence of Definitions of Real Natural Logarithm

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Theorem

The following definitions of the concept of Real Natural Logarithm are equivalent:

Definition 1

Let $x \in \R$ be a real number such that $x > 0$.

The (natural) logarithm of $x$ is defined as:

$\displaystyle \ln x := \int_1^x \frac {\d t} t$

Definition 2

Let $x \in \R$ be a real number such that $x > 0$.

The (natural) logarithm of $x$ is defined as:

$\ln x := y \in \R: e^y = x$

where $e$ is Euler's number.

Definition 3

Let $x \in \R$ be a real number such that $x > 0$.

The (natural) logarithm of $x$ is defined as:

$\ds \ln x := \lim_{n \mathop \to \infty} n \paren {\sqrt [n] x - 1}$


Proof

Definition 1 implies Definition 2

Let $\map F x$ be $\ds \int_1^x \frac {\d t} t$.

Let $\map f t$ be $\ds \int \frac {\d t} t$.

Then:

$\dfrac {\d t} t = \dfrac 1 t$

Or:

$\dfrac {\d x} x = \dfrac 1 x$

Also:

$\map F x = \map f x - \map f 1$

Therefore:

\(\ds \frac {\d \map F x} {\d x}\) \(=\) \(\ds \frac {\d \map f x} {\d x} - \frac {\d \map F 1} {\d x}\)
\(\ds \) \(=\) \(\ds \frac {\d \map f x} {\d x}\) Derivative of Constant
\(\ds \) \(=\) \(\ds \frac 1 x\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d x} {\d \map F x}\) \(=\) \(\ds x\) Derivative of Inverse Function

Furthermore:

$\map F 1 = \map f 1 - \map f 1 = 0$



The result follows from the fifth definition of the exponential function:

$\map F x \equiv e^x$

$\Box$


Definition 2 implies Definition 1

\(\ds e^{\map F x}\) \(=\) \(\ds x\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d x} {\d \map F x}\) \(=\) \(\ds x\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d \map F x} {\d x}\) \(=\) \(\ds \frac 1 x\) Derivative of Inverse Function

Let $\map f t$ be $\ds \int \frac 1 t \rd t$.

Then:

$\map F x = \map f x + C$

When $\map F x = 0$:

$x = e^{\map F x} = 1$
$\map F 1 = \map f 1 + C = 0 \implies \map f 1 = - C$

Therefore:

$\map F x = \map f x - \map f 1$

Therefore:

$\ds \map F x = \int_1^x \frac {\d t} t$

$\Box$


Therefore:

\(\ds y = \int_1^x \frac {\d t} t\) \(\leadsto\) \(\ds e^y = x\) Definition 1 implies Definition 2
\(\ds e^y = x\) \(\leadsto\) \(\ds y = \int_1^x \frac {\d t} t\) Definition 2 implies Definition 1
\(\ds \leadsto \ \ \) \(\ds y = \int_1^x \frac {\d t} t\) \(\iff\) \(\ds e^y = x\)

$\blacksquare$