Potenzgesetze

$2^3\cdot 2^2=2^{3+2}$ $$ $\begin{array}{l}=(2\cdot 2\cdot 2)\cdot (2\cdot 2)\\ =2\cdot 2\cdot 2\cdot 2\cdot 2=2^5\end{array}$

$a^x\cdot a^y=a^{x+y}$

Multiplikation bei $\begin{array}{l}\\ \\ \end{array}$gleicher Basis $a$

$\frac{2^3}{2^2}=2^{3-2}$ $$ $\begin{array}{l}=\frac{2\cdot 2\cdot 2}{2\cdot 2}\\ =2^1=2^{3-2}\end{array}$

$\frac{a^x}{a^y}=a^{x-y}$

Division bei $\begin{array}{l}\\ \\ \end{array}$gleicher Basis $a$

$2^3\cdot 3^3={(2\cdot 3)}^3$ $$ $\begin{array}{l}=(2\cdot 2\cdot 2)\cdot (3\cdot 3\cdot 3)\\ =(2\cdot 3)\cdot (2\cdot 3)\cdot (2\cdot 3)\\ =(2\cdot 3{)}^3\end{array}$

$a^x\cdot b^x={(a\cdot b)}^x$

Multiplikation bei $\begin{array}{l}\\ \\ \end{array}$gleichem Exponenten $x$

$\frac{2^3}{3^3}={\left(\frac{2}{3}\right)}^3$ $$ $\begin{array}{l}=\frac{2\cdot 2\cdot 2}{3\cdot 3\cdot 3}\\ =\frac{2}{3}\cdot \frac{2}{3}\cdot \frac{2}{3}\\ ={\left(\frac{2}{3}\right)}^3\end{array}$

$\frac{a^x}{b^x}={\left(\frac{a}{b}\right)}^x$

Division bei $\begin{array}{l}\\ \\ \end{array}$gleichem Exponenten $x$

${(-2)}^6=2^6$ $$ $\begin{array}{l}=(-2{)}^2\cdot (-2{)}^2\cdot (-2{)}^2\\ =2^2\cdot 2^2\cdot 2^2\\ =2^6\end{array}$

${(-a)}^x=a^x$

Negative Basis und $\begin{array}{l}\\ \\ \end{array}$gerader Exponent.

${(-2)}^5=-(2^5)$ $$ $\begin{array}{l}=(-2{)}^2\cdot (-2{)}^2\cdot (-2)\\ =2^2\cdot 2^2\cdot (-2)\\ =-(2^5)\end{array}$

$\text{}{(-a)}^x=-(a^x)$

Negative Basis und $\begin{array}{l}\\ \\ \end{array}$ungerader Exponent.

$2^0=1$ $$ $\begin{array}{l}{2}^2=2\cdot 2=2\cdot 2\cdot 1\\ 2^1=2=2\cdot 1\\ 2^0=1\end{array}$

$a^0=1$

Null im Exponenten $\begin{array}{l}\\ \\ \end{array}$bei beliebiger Basis $a\ne 0$

$2^{-3}=\frac{1}{2^3}$ $$ $\begin{array}{l}1=2^0=2^{3+(-3)}\\ =2^3\cdot 2^{-3}\\ \end{array}$ Damit dies $1$ ist, muss $$ $\begin{array}{l}{2}^{-3}=\frac{1}{2^3}\\ \end{array}$ gelten, denn $$ $2^3\cdot \frac{1}{2^3}=\frac{2^3}{2^3}=1$

$a^{-x}=\frac{1}{a^x}$

Negative Exponenten

$2^{\frac{1}{3}}=\sqrt[3]{2}$ $\begin{array}{l}\\ \\ \end{array}$ $\begin{array}{l}2=2^1=2^{\frac{1}{3}\cdot 3}\\ ={\left(2^{\frac{1}{3}}\right)}^3\\ \end{array}$ Damit dies $2$ ist, muss $\begin{array}{l}\\ \\ \end{array}$ $2^{\frac{1}{3}}=\sqrt[3]{2}$ gelten, $\begin{array}{l}\\ \\ \end{array}$ denn ${\left(\sqrt[3]{2}\right)}^3=2$

$a^{\frac{1}{n}}=\sqrt[n]{a}$

Einheitsbrüche $\begin{array}{l}\\ \\ \end{array}$im Exponenten