Solution 1.3:5b

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Current revision (14:24, 22 September 2008) (edit) (undo)
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If we use
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If we use <math>4 = 2\cdot 2 = 2^{2}</math>, the power rules give
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<math>4=2\centerdot 2=2^{2}</math>, the power rules give
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{{Displayed math||<math>4^{-\frac{1}{2}} = \bigl( 2^{2}\bigr)^{-\frac{1}{2}} = 2^{2\cdot (-\frac{1}{2})} = 2^{-1} = \frac{1}{2}\,</math>.}}
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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>
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Current revision

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

\displaystyle 4^{-\frac{1}{2}} = \bigl( 2^{2}\bigr)^{-\frac{1}{2}} = 2^{2\cdot (-\frac{1}{2})} = 2^{-1} = \frac{1}{2}\,.
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