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April 18, 2014

April 18, 2014

Posted by **Carmin** on Thursday, June 25, 2009 at 12:55am.

(root5 a^7/2)^4 a^3/a^8

So confused, I tried figuring this out but need some help tracing the steps. I would appreciate some feedback! Thanks!

- Algebra -
**MathMate**, Thursday, June 25, 2009 at 7:16amI have assumed that the fraction 7/2 is enclosed in parentheses, as follows:

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

Since the bases of the exponents are "a" or 5, you only have to apply the following rules of exponents to get your answer of 25a^{9}.

1. sqrt(5) = 5^{1/2}

2. distributive law:

(ab)^{x}= a^{x}b^{x}

3. addition of exponents of same base

a^{x}a^{y}= a^{x+y}

4. multiplication of exponents

(a^{x})^{y}= a^{xy}

So:

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

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

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

=25 a^{9}

- Algebra -
**Carmin**, Thursday, June 25, 2009 at 7:55amWOW Thanks for the reply, but I still do not get how we get 25a^9. Can you please further explain. Thanks!

- Algebra -
**Carmin**, Thursday, June 25, 2009 at 8:14amI get 25a^9/16? How can this be? Please help, still confused, thanks!

- Algebra -
**MathMate**, Thursday, June 25, 2009 at 8:37amIf you show your steps, it may help me spot where it has gone wrong.

- Algebra -
**MathMate**, Thursday, June 25, 2009 at 6:34pm...

=(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.

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