(related to Proposition: Natural Logarithm)
Let \(n\in\mathbb N\). From the definition of the exponential function, it follows that
\[\exp(n)=1+n+\frac {n^2}2+\ldots \ge 1+n.\quad \quad ( * )\]
Due to the reciprocity of exponential function, we get with \( ( * ) \) the approximation
\[\exp(-n)=\frac 1{\exp(n)}\le \frac 1{1+n}.\quad\quad ( * * )\]
Since the exponential function is strictly monotonically increasing and is continuous, we can conclude from the lemma about invertible functions that the it is invertible on closed real intervals \(\left[-n,n\right]\) for all \(n\in\mathbb N\). Note that \(\bigcup_{n\in\mathbb N}[-n,n]=\mathbb R\) and \(\bigcup_{n\in\mathbb N}\left[\frac 1{n+1},n+1\right]=\mathbb R_+^*\), where \(\mathbb R_+^*\) denotes the set of all positive real numbers. Therefore, it follows that the natural logarithm
\[\ln:\mathbb R_{+}^*\to\mathbb R\]
is the inverse function of the exponential function and that it is also continuous, strictly monotonically increasing.