Lösung 1.3:5e

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Individually, <math>3^{1\textrm{.}4}</math> and <math>3^{0\textrm{.}6}</math> are difficult to calculate, but when multiplied together
Individually, <math>3^{1\textrm{.}4}</math> and <math>3^{0\textrm{.}6}</math> are difficult to calculate, but when multiplied together
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{{Displayed math||<math>3^{1\textrm{.}4}\cdot 3^{0\textrm{.}6} = 3^{1\textrm{.}4+0\textrm{.}6} = 3^{2} = 3\cdot 3 = 9</math>}}
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{{Abgesetzte Formel||<math>3^{1\textrm{.}4}\cdot 3^{0\textrm{.}6} = 3^{1\textrm{.}4+0\textrm{.}6} = 3^{2} = 3\cdot 3 = 9</math>}}
on using the power rules.
on using the power rules.

Version vom 08:19, 22. Okt. 2008

Individually, \displaystyle 3^{1\textrm{.}4} and \displaystyle 3^{0\textrm{.}6} are difficult to calculate, but when multiplied together

\displaystyle 3^{1\textrm{.}4}\cdot 3^{0\textrm{.}6} = 3^{1\textrm{.}4+0\textrm{.}6} = 3^{2} = 3\cdot 3 = 9

on using the power rules.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_1.3:5e“