Natural Logarithm of 1 is 0

Theorem

$\ln 1 = 0$

where $\ln 1$ denotes the natural logarithm of $1$.

Proof 1

We use the definition of the natural logarithm as an integral:

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

From Integral on Zero Interval:

$\ds \ln 1 = \int_1^1 \frac {\d t} t = 0$

$\blacksquare$

Proof 2

We use the definition of the natural logarithm as the inverse of the exponential:

$\ln x = y \iff e^y = x$

Then:

\(\ds e^0\)

\(=\)

\(\ds 1\)

Exponential of Zero

\(\ds \leadstoandfrom \ \ \)

\(\ds \ln 1\)

\(=\)

\(\ds 0\)

$\blacksquare$

Proof 3

We use the definition of the natural logarithm as the limit of a sequence:

$\ds \ln x = \lim_{n \mathop \to \infty} n \paren {\sqrt [n] x - 1}$

Then:

\(\ds \ln 1\)

\(=\)

\(\ds \lim_{n \mathop \to \infty} n \paren {\sqrt [n] 1 - 1}\)

\(\ds \)

\(=\)

\(\ds \lim_{n \mathop \to \infty} n \times 0\)

\(\ds \)

\(=\)

\(\ds \lim_{n \mathop \to \infty} 0\)

\(\ds \)

\(=\)

\(\ds 0\)

$\blacksquare$