Natural Logarithm of 1 is 0

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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:

\(\ds \ln 1\) \(=\) \(\ds \lim_{n \mathop \to \infty} n \paren {\sqrt [n] 1 - 1}\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} n \times 0\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} 0\)
\(\ds \) \(=\) \(\ds 0\)

$\blacksquare$