Logarithms/Exponential functions/Introduction/Section

The exponential functions for various bases

Definition

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

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

Theorem

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

$f\colon \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto b^{x},$

fulfills the following properties.

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

Proof

See exercise.

$\Box$

Remark

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

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

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 ${}x$, one takes a sequence ${}q_{n}$ of rational numbers converging to ${}x$, and defines

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

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

Definition

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

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

Logarithms for various bases

Theorem

The logarithms to base

${}b$ fulfill the following rules.

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

   ${}\log _{a}y=\log _{a}{\left(b^{\log _{b}y}\right)}=\log _{b}y\cdot \log _{a}b\,.$

Proof

See exercise.

$\Box$