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:
- $\map F 1 = \map f 1 - \map f 1 = 0$
Why is the reverse of $\map F x$ a function?
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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 | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y = \int_1^x \frac {\d t} t\) | \(\iff\) | \(\ds e^y = x\) |
$\blacksquare$
Add definition 3
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