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August 1, 2014

August 1, 2014

Posted by **Angie** on Thursday, June 25, 2009 at 5:48pm.

(root5a^7/2)^4 a^3/a^8

- Math -
**MathMate**, Thursday, June 25, 2009 at 6:20pmThe problem has been solved before. See following link for explanations.

http://www.jiskha.com/display.cgi?id=1245905705

- Math -
**MathMate**, Thursday, June 25, 2009 at 6:35pmI have added the following clarification for Carmin. It is reproduced here in case you are also interested.

...

=(sqrt(5)^{4}*(a^{(7/2)})^{4}) a^{3-8}

=5^{(1/2)*4}*a^{(7/2)*4}*a^{3-8}

=25 a^{14}a^{-5}

=25 a^{9}

the denominator 2 in the fraction (7/2) is an exponent and not a simple number. This is probably from where you got the 16.

- Math -
**Angie**, Thursday, June 25, 2009 at 6:46pmSo the 16 is not part of the answer? It's just 25 a^9?

- Math -
**MathMate**, Thursday, June 25, 2009 at 8:31pmRight!

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