# Logarithm Function

The inverse of the exponential function.

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Concept | Content |
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Exponential Function | A special function that can be defined via a power series. |

Injectivity, Surjectivity, Bijectivity | These are important notions for maps. |

Continuity | The concept that relates functions with convergent sequences. |

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## Study *Logarithm Function* #

We will now define the logarithm as the inverse function of the exponential function. As we have seen, the exponential function is bijective as a map from \(\mathbb{R}\) to \((0,\infty)\). This justifies the following definition.

**Definition 1**. *The (natural) logarithm
\(\log:(0,\infty)\to\mathbb{R}\) is
defined as the inverse function of \(\exp:\mathbb{R}\to(0,\infty)\), i.e., for
all \(x\in\mathbb{R}\) holds \(\log(\exp(x))=x\) and for all \(y \in (0,\infty)\) holds \(\exp(\log(y)) = y\).*

In many books, the above defined function is also denoted by
*logarithmus naturalis* \(\ln\). Before we collect some properties,
a general result about continuity of inverse functions is presented.

**Theorem 2** (Continuity of the inverse function).
*Let \(I,J\subset\mathbb{R}\) be
open intervals and let \(f:I\to J\) be
a continuous, bijective and strictly monotonically increasing (or
decreasing) function. Then the inverse function \(f^{-1}:J\to I\) is continuous and strictly
monotonically increasing (resp. decreasing).*

*Proof:* First we show strict monotonic increase.
Let \(y_1,y_2\in J\) with \(y_1< y_2\). Then \[f(f^{-1}(y_1))=y_1 < y_2 =
f(f^{-1}(y_2))\] shows that, since \(f\) is strictly monotonically increasing,
\(f^{-1}(y_1)\geq f^{-1}(y_2)\) cannot
hold, so that \(f^{-1}(y_1) <
f^{-1}(y_2)\). But this means that \(f^{-1}\) is strictly monotonically
increasing.

Now we show continuity. Let \(\varepsilon>0\) and \(y_0\in J=f(I)\). Set \(x_0:=f^{-1}(y_0)\in I\). Since \(I\) is an open interval, there is an \(\varepsilon'>0\) with \(\varepsilon'<\varepsilon\) such that
\([x_0-\varepsilon',x_0+\varepsilon']\subset
I\). Since \(f\) is strictly
monotonically increasing, \[\delta:=\min\{f(x_0+\varepsilon')-y_0,y_0-f(x_0-\varepsilon')\}>0.\]
Then for \(y\in J\) with \(|y-y_0|<\delta\) holds \(f(x_0-\varepsilon')< y<
f(x_0+\varepsilon')\) and the intermediate value theorem
yields \[x:=f^{-1}(y) \in
[x_0-\varepsilon',x_0+\varepsilon'] \subset ~
(x_0-\varepsilon,x_0+\varepsilon) ~ = ~
(f^{-1}(y_0)-\varepsilon,f^{-1}(y_0)+\varepsilon)\ ,\] i.e. \(|f^{-1}(y)-f^{-1}(y_0)|<\varepsilon\).
By the \(\varepsilon\)-\(\delta\) criterion this means that \(f^{-1}\) is continuous in \(y_0\). \(\Box\)

We want to remark that Theorem 2 also holds
for intervals \(I\) and \(J\) which are not open. In this case the
proof of continuity of \(f^{-1}\) in
\(y_0\in J\) has to be slightly adapted
for boundary points \(x_0:=f^{-1}(y_0)\). This was dropped for
simplicity reasons.

**Theorem 3** (Properties of the Logarithm).

*For all \(x,y\in~(0,\infty)\) holds \(\log(x\cdot y)=\log(x)+\log(y)\).**\(\log:~(0,\infty)~\to~\mathbb{R}\) is strictly monotonically increasing.**\(\log:~(0,\infty)~\to~\mathbb{R}\) is continuous.*

Fabian Gabel and Marcus Waurick. *Well-defined & Wonderful podcast*, marcus-waurick.de.

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