# 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:

$\displaystyle \ln x := \lim_{n \to \infty} n \left({ \sqrt[n]{ x } - 1 }\right)$

## Proof

### Definition 1 implies Definition 2

Let $F \left({x}\right)$ be $\displaystyle \int_1^x \frac {\d t} t$.

Let $f \left({t}\right)$ be $\displaystyle \int \frac {\d t} t$.

Then:

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

Or:

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

Also:

$F \left({x}\right) = f \left({x}\right) - f \left({1}\right)$

Therefore:

 $\displaystyle \frac {\d F \left({x}\right)} {\d x}$ $=$ $\displaystyle \frac {\d f \left({x}\right)} {\d x} - \frac {\d F \left({1}\right)} {\d x}$ $\displaystyle$ $=$ $\displaystyle \frac {\d f \left({x}\right)} {\d x}$ Derivative of Constant $\displaystyle$ $=$ $\displaystyle \frac 1 x$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d x} {\d F \left({x}\right)}$ $=$ $\displaystyle x$ Derivative of Inverse Function

Furthermore:

$F \left({1}\right) = f \left({1}\right) - f \left({1}\right) = 0$

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

$F \left({x}\right) \equiv e^x$

$\Box$

### Definition 2 implies Definition 1

 $\displaystyle e^{F \left({x}\right)}$ $=$ $\displaystyle x$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d x} {\d F \left({x}\right)}$ $=$ $\displaystyle x$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d F \left({x}\right)} {\d x}$ $=$ $\displaystyle \frac 1 x$ Derivative of Inverse Function

Let $f \left({t}\right)$ be $\displaystyle \int \frac 1 t \rd t$.

Then:

$F \left({x}\right) = f \left({x}\right) + C$

When $F \left({x}\right) = 0$:

$x = e^{F \left({x}\right)} = 1$
$F \left({1}\right) = f \left({1}\right) + C = 0 \implies f \left({1}\right) = - C$

Therefore:

$F \left({x}\right) = f \left({x}\right) - f \left({1}\right)$

Therefore:

$\displaystyle F \left({x}\right) = \int_1^x \frac {\d t} t$

$\Box$

Therefore:

 $\displaystyle y = \int_1^x \frac {\d t} t$ $\leadsto$ $\displaystyle e^y = x$ Definition 1 implies Definition 2 $\displaystyle e^y = x$ $\leadsto$ $\displaystyle y = \int_1^x \frac {\d t} t$ Definition 2 implies Definition 1 $\displaystyle \leadsto \ \$ $\displaystyle y = \int_1^x \frac {\d t} t$ $\iff$ $\displaystyle e^y = x$

$\blacksquare$