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anonymous
 5 years ago
2 raised to the power of 5/2  2 raised to the power of 3/2
anonymous
 5 years ago
2 raised to the power of 5/2  2 raised to the power of 3/2

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When you raise a power to a power you multiply the exponents.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well my answers are either A. 2 1/2 B. 2 C. 2 3/2 D. 2 5/3 or E. 2^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, what is (5/22)(3/2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[(2^{\frac{5}{2}2})^{\frac{3}{2}}\] \[= 2^{\frac{3}{2}(\frac{5}{2}2)}\] \[= 2^{\frac{3}{2}\frac{5}{2}  \frac{3}{2}\frac{2}{1}}\] \[= 2^{\frac{15}{4}  \frac{12}{4}}\] \[= 2^{\frac{3}{4}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Wait. I think you have a different problem.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats not one of the answers ,the question is 2 raised to 5 over 2 subtract 2 raised to 3 over 2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, you should put parens on that.. Do you mean \[2^{\frac{5}{2}}  2^{\frac{3}{2}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Very different problem ;p

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can factor a 1/2 from the exponents.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually no you cannot.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That's just \(\sqrt{32}  \sqrt{8}\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if i do 2 5/2 divided by half i get 9 and if i do 2 3/2 divided by half i get 7 so if i subtract them i obviously get 2 which is b but i dont think thats right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\(\sqrt{32} = \sqrt{2^2*2^2*2} = 2*2\sqrt{2} = 4\sqrt{2}\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the answer is 2 and 3/2 but i dont no how they got that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\(\sqrt{8} = \sqrt{2^2*2} = 2\sqrt{2}\) \[\implies \sqrt{32}  \sqrt{8} = 4\sqrt{2}  2\sqrt{2} = 2\sqrt{2}\] So the answer should be \(2^{\frac{3}{2}}\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where are you getting the square roots from

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\(2^{\frac{1}{2}} = \sqrt{2}\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well think about it this way then.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Multiplying powers of the same base you add their exponents. \(2^{5/2} \) \(= 2^{4/2} \bullet 2^{1/2}\) \(=2^2 \bullet 2^{1/2}\) \(= 4(2^{1/2})\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Same for \(2^{3/2} = 2(2^{1/2})\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So \(4(2^{1/2})  2(2^{1/2}) = 2(2^{1/2})\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\(= 2^1* 2^{1/2} = 2^{1+1/2} = 2^{3/2}\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so i just add 5/2 + 3/2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no. Watch this series on exponents.. http://www.khanacademy.org/video/exponentproperties1?playlist=Developmental%20Math

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha i did but i dont get subtraction... with exponents

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That's because you can't do anything special when you are adding and subtracting exponents. You can only factor out common factors. But in this case you have a common factor of 2^{1/2} that you can pull out of each term.
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