Lösung 1.3:5b

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If we use
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<center> [[Image:1_3_5b.gif]] </center>
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<math>4=2\centerdot 2=2^{2}</math>, the power rules give
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<math>4^{-\frac{1}{2}}=\left( 2^{2} \right)^{-\frac{1}{2}}=2^{2\centerdot \left( -\frac{1}{2} \right)}=2^{-1}=\frac{1}{2}</math>

Version vom 12:01, 15. Sep. 2008

If we use \displaystyle 4=2\centerdot 2=2^{2}, the power rules give


\displaystyle 4^{-\frac{1}{2}}=\left( 2^{2} \right)^{-\frac{1}{2}}=2^{2\centerdot \left( -\frac{1}{2} \right)}=2^{-1}=\frac{1}{2}

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