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:

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

$\blacksquare$