# Algebra

posted by on .

(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 - ,

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

• Algebra - ,

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

• Algebra - ,

• Algebra - ,

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

• Algebra - ,

...
=(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.

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