Equivalence of Definitions of Real Natural Logarithm

Definition 1 implies Definition 2
Let $F \left({x}\right)$ be $\displaystyle \int_1^x \frac {\mathrm d t} t$.

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

Then:
 * $\dfrac {\mathrm d t} t = \dfrac 1 t$

Or:
 * $\dfrac {\mathrm d x} x = \dfrac 1 x$

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

Therefore:

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

Definition 2 implies Definition 1
Let $f \left({t}\right)$ be $\displaystyle \int \frac 1 t \ \mathrm d t$.

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

Therefore: