# Algebra

(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!

1. I have assumed that the fraction 7/2 is enclosed in parentheses, as follows:

(sqrt(5)*a(7/2))4 a3/a8

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 25a9.

1. sqrt(5) = 51/2
2. distributive law:
(ab)x = ax bx
3. addition of exponents of same base
ax ay = ax+y
4. multiplication of exponents
(ax )y = axy

So:
(sqrt(5)*a(7/2))4 a3/a8
=(sqrt(5)4*(a(7/2))4) a3-8
=25 a14 a-5
=25 a9

posted by MathMate
2. WOW Thanks for the reply, but I still do not get how we get 25a^9. Can you please further explain. Thanks!

posted by Carmin

posted by Carmin
4. If you show your steps, it may help me spot where it has gone wrong.

posted by MathMate
5. ...
=(sqrt(5)4*(a(7/2))4) a3-8
=5(1/2)*4*a(7/2)*4*a3-8
=25 a14 a-5
=25 a9

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.

posted by MathMate

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