# Equivalence of Definitions of Real Natural Logarithm

## 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 | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds y = \int_1^x \frac {\d t} t\) | \(\iff\) | \(\ds e^y = x\) |

$\blacksquare$