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 August 10th, 2009, 10:49 AM #1 Newbie   Joined: Mar 2009 Posts: 24 Thanks: 0 fractional exponents Solve the problem: 2^(5/2)-2^(3/2) I'm really stuck on this problem, can someone explain how to go about getting the answer? Thank you!
 August 10th, 2009, 11:24 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,931 Thanks: 1125 Math Focus: Elementary mathematics and beyond Re: fractional exponents $2^{\frac{5}{2}}-2^{\frac{3}{2}}=\sqrt{2^5}-\sqrt{2^3}=\sqrt{32}-\sqrt{8}=\sqrt{4}\sqrt{8}-\sqrt{8}=2\sqrt{8}-\sqrt{8}=\sqrt{8}=2\sqrt{2}$
 August 10th, 2009, 12:18 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,471 Thanks: 2038 $\text{Or }2^{\small{\frac52}}\,-\,2^{\small{\frac32}}\,=\,(2^{\small2}\,-\,2^{\small1})2^{\small{\frac12}}\,=\,2\sqrt2.$
 August 10th, 2009, 03:56 PM #4 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Let $k=2^{\frac{5}{2}}-2^{\frac{3}{2}}\Rightarrow \frac{k}{2^{\frac{3}{2}}}=\frac{2^{\frac{5}{2}}-2^{\frac{3}{2}}}{2^{\frac{3}{2}}}=2^{\frac{2}{2}}-2^0=2-1=1\Rightarrow\therefore k=(2^{\frac{3}{2}})(1)=2\sqrt{2}.$

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