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:

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

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:

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

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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

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Furthermore:

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

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

$\map F x \equiv e^x$

There is believed to be a mistake here, possibly a typo. In particular: But it was supposed to be $\map F x = \ln x$. You can help ProofWiki by reviewing it, and either correcting it or adding some explanatory material as to why you believe it is actually correct after all. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Mistake}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

$\Box$

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

\(\ds \leadsto \ \ \)

\(\ds y = \int_1^x \frac {\d t} t\)

\(\iff\)

\(\ds e^y = x\)

$\blacksquare$

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