Potenser og rotuttrykk

$\sqrt[3]{8}=2$ fordi $2^3=8$

eller

$\sqrt[3]{8}=8^{\frac{1}{3}}={\left(2^3\right)}^{\frac{1}{3}}=2^{3·\frac{1}{3}}=2^1=2$

$\sqrt[3]{27}=3$ fordi $3^3=27$

eller

$\sqrt[3]{27}={27}^{\frac{1}{3}}={\left(3^3\right)}^{\frac{1}{3}}=3^{3·\frac{1}{3}}=3^1=3$

$\sqrt[4]{81}=3$ fordi $3^4=81$

$$ \sqrt[4]{81} = 3 = {81}^{\frac{1}{4 }}= {\left(3^4\right)}^{\frac{1}{4 }}= 3^{4\frac{1}{4 }}= 3^1 = 3$$

$$ \sqrt[5]{15,25} = 1,72$$

$$ \sqrt[8]{100} = 1,78$$

$$ \sqrt[9]{2,25} = 1,09$$

$$ 9^{\frac{1}{2}} = \sqrt{9 }= 3$$

$$ {27}^{\frac{1}{3}} = \sqrt[3]{27 }= 3$$

$$ \sqrt[5]{32} = 2$$

$$ {256}^{\frac{1}{4}} = \sqrt[4]{256} = 4$$

$$ \sqrt[3]{125} = 5$$

$$ \sqrt[3]{9}·\sqrt[3]{3} =\sqrt[ 3]{9·3} = \sqrt[3]{27 }= 3$$

$$ \frac{\sqrt[3]{54}}{\sqrt[3]{2}} = \sqrt[3]{\frac{54}{2}} = \sqrt[3]{27 }= 3$$

$\begin{array}{c}\begin{array}{l}{8}^{\frac{2}{3}}={\left(2^3\right)}^{\frac{2}{3}}=2^{3\frac{2}{3}}=2^2=4\\ \\ 8^{\frac{2}{3}}=\sqrt[3]{8^2}=\sqrt[3]{64}=4\end{array}\end{array}$

$6^{\frac{2}{3}}=\sqrt[3]{6^2}=\sqrt[3]{36}$

$4^{\frac{3}{2}}=\sqrt{4^3}=4\sqrt{4}=4·2=8$

$9^{\frac{3}{2}}=\sqrt{9^3}=9\sqrt{9}=9·3=27$

$3^{\frac{4}{3}}=3^{1+\frac{1}{3}}=3^1·3^{\frac{1}{3}}=3·\sqrt[3]{3}$

$5^{\frac{3}{5}}=5^{1+\frac{1}{3}}=5^1·5^{\frac{2}{3}}=5\sqrt[3]{5^2}=5·\sqrt[3]{25}$

$2^{\frac{7}{3}}=2^{2+\frac{1}{3}}=2^2·2^{\frac{1}{3}}=4·\sqrt[3]{2}$

$$ {27}^{-\frac{1}{3}} = \frac{1}{{27}^{{\displaystyle \frac{1}{3}}}} = \frac{1}{\sqrt[3]{27}} = \frac{1}{3}$$

$$ 2^{\frac{2}{3}}·2^{\frac{1}{3}} = 2^{\frac{2}{3}+\frac{1}{3}}= 2^{\frac{3}{3}}= 2$$

$$ 4^{\frac{5}{2}}·4^{-\frac{3}{2}} = 4^{\frac{5}{2}-\frac{3}{2}}= 4^{\frac{2}{2}}= 4$$

$$ \begin{array}{rcl}{2}^{\frac{5}{2}}·4^{-\frac{3}{2}}& =& 2^{\frac{5}{2}}·{\left(2^2\right)}^{-\frac{3}{2}}= 2^{\frac{5}{2}-\frac{6}{2}}\\ & =& 2^{-\frac{1}{2}}=\frac{1}{\sqrt{2}} = \frac{1·\sqrt{2}}{\sqrt{2}·\sqrt{2}} = \frac{\sqrt{2}}{2}\end{array}$$

$$ {\left(3^4\right)}^{\frac{1}{4}} = 3^{4·\frac{1}{4}} = 3^{\frac{4}{4}} = 3$$

$$ \begin{array}{rcl}{\left(3^4\right)}^{\frac{1}{4}} & =& 3^{\frac{4}{3}+\frac{1}{3}-\left(-\frac{1}{3}\right)} = 3^{\frac{4}{3}+\frac{1}{3}+\frac{1}{3}}\\ & =& 3^{\frac{6}{3}}= 3^2 = 9\end{array}$$

$$ \frac{4^{{\displaystyle \frac{4}{3}}}·2^{{\displaystyle \frac{1}{3}}}}{2^2} = \frac{2^{{\displaystyle \frac{8}{3}}}·2^{{\displaystyle \frac{1}{3}}}}{2^2} = 2^{\frac{8}{3}+\frac{1}{3}-2} = 2^1 = 2$$

Vi må løse likningen $\begin{array}{rcl}4\pi r^2& = & 17\end{array}$.

Radien er 1,16 cm.

Vi må løse likningen $\begin{array}{rcl}\frac{4\pi r^3}{3}& = & 9,35\end{array}$

Radien er 1,31 cm.