Posted by Carmin on Thursday, June 25, 2009 at 12:55am.
Please Simplify:
(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 7:16am
I 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^{38}
=25 a^{14} a^{5}
=25 a^{9}

Algebra  Carmin, Thursday, June 25, 2009 at 7:55am
WOW 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 8:14am
I get 25a^9/16? How can this be? Please help, still confused, thanks!

Algebra  MathMate, Thursday, June 25, 2009 at 8:37am
If 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^{38}
=5^{(1/2)*4}*a^{(7/2)*4}*a^{38}
=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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