Lösung 1.3:5c

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We can write
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<center> [[Image:1_3_5c.gif]] </center>
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<math>9=3\centerdot 3=3^{2}</math>
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and then the power rules give
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<math>9^{\frac{3}{2}}=\left( 3^{2} \right)^{\frac{3}{2}}=3^{2\centerdot \frac{3}{2}}=3^{3}=3\centerdot 3\centerdot 3=27</math>

Version vom 12:03, 15. Sep. 2008

We can write \displaystyle 9=3\centerdot 3=3^{2} and then the power rules give


\displaystyle 9^{\frac{3}{2}}=\left( 3^{2} \right)^{\frac{3}{2}}=3^{2\centerdot \frac{3}{2}}=3^{3}=3\centerdot 3\centerdot 3=27

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