# Logarithms/Exponential functions/Introduction/Section

## Definition

For a positive real number ${\displaystyle {}b>0}$, the exponential function for the base ${\displaystyle {}b}$ is defined as

${\displaystyle {}b^{x}:=\exp(x\ln b)\,.}$

## Theorem

Let ${\displaystyle {}b}$ denote a positive real number. Then the exponential function

${\displaystyle f\colon \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto b^{x},}$
fulfills the following properties.
1. We have ${\displaystyle {}b^{x+x'}=b^{x}\cdot b^{x'}}$ for all ${\displaystyle {}x,x'\in \mathbb {R} }$.
2. We have ${\displaystyle {}b^{-x}={\frac {1}{b^{x}}}}$.
3. For ${\displaystyle {}b>1}$ and ${\displaystyle {}x>0}$, we have ${\displaystyle {}b^{x}>1}$.
4. For ${\displaystyle {}b<1}$ and ${\displaystyle {}x>0}$, we have ${\displaystyle {}b^{x}<1}$.
5. For ${\displaystyle {}b>1}$, the function ${\displaystyle {}f}$ is strictly increasing.
6. For ${\displaystyle {}b<1}$, the function ${\displaystyle {}f}$ is strictly decreasing.
7. We have ${\displaystyle {}(b^{x})^{x'}=b^{x\cdot x'}}$ for all ${\displaystyle {}x,x'\in \mathbb {R} }$.
8. For ${\displaystyle {}a\in \mathbb {R} _{+}}$, we have ${\displaystyle {}(ab)^{x}=a^{x}\cdot b^{x}}$.

### Proof

${\displaystyle \Box }$

## Remark

There is another way to introduce the exponential function ${\displaystyle {}x\mapsto a^{x}}$ to base ${\displaystyle {}a>0}$. For a natural number ${\displaystyle {}n\neq 0}$, one takes the ${\displaystyle {}n}$th product of ${\displaystyle {}a}$ with itself as definition for ${\displaystyle {}a^{n}}$. For a negative integer ${\displaystyle {}x}$, one sets ${\displaystyle {}a^{x}:=(a^{-x})^{-1}}$. For a positive rational number ${\displaystyle {}x=r/s}$, one sets

${\displaystyle {}a^{x}:={\sqrt[{s}]{a^{r}}}\,,}$

where one has to show that this is independent of the chosen representation as a fraction. For a negative rational number, one takes again the inverse. For an arbitrary real number ${\displaystyle {}x}$, one takes a sequence ${\displaystyle {}q_{n}}$ of rational numbers converging to ${\displaystyle {}x}$, and defines

${\displaystyle {}a^{x}:=\lim _{n\rightarrow \infty }a^{q_{n}}\,.}$

For this, one has to show that these limits exist and that they are independent of the chosen rational sequence. For the passage from ${\displaystyle {}\mathbb {Q} }$ to ${\displaystyle {}\mathbb {R} }$, the concept of uniform continuity is crucial.

## Definition

For a positive real number ${\displaystyle {}b>0}$, ${\displaystyle {}b\neq 1}$, the logarithm to base ${\displaystyle {}b}$ of ${\displaystyle {}x\in \mathbb {R} _{+}}$ is defined by

${\displaystyle {}\log _{b}x:={\frac {\ln x}{\ln b}}\,.}$

## Theorem

${\displaystyle {}b}$ fulfill the following rules.
1. We have ${\displaystyle {}\log _{b}(b^{x})=x}$ and ${\displaystyle {}b^{\log _{b}(y)}=y}$, this means that the logarithm to Base ${\displaystyle {}b}$ is the inverse function for the exponential function to base ${\displaystyle {}b}$.
2. We have ${\displaystyle {}\log _{b}(y\cdot z)=\log _{b}y+\log _{b}z}$.
3. We have ${\displaystyle {}\log _{b}y^{u}=u\cdot \log _{b}y}$ for ${\displaystyle {}u\in \mathbb {R} }$.
4. We have
${\displaystyle {}\log _{a}y=\log _{a}{\left(b^{\log _{b}y}\right)}=\log _{b}y\cdot \log _{a}b\,.}$

### Proof

${\displaystyle \Box }$