SOLUTION: 4a^-5b^5 -------- 12a^3b^-5 all to the 0 power.. would this just be 1? I know it would simplify to 3b^10 but i dont know what to do with the 0 power

Algebra ->  Exponents -> SOLUTION: 4a^-5b^5 -------- 12a^3b^-5 all to the 0 power.. would this just be 1? I know it would simplify to 3b^10 but i dont know what to do with the 0 power       Log On


   

Question 466787: 4a^-5b^5
--------
12a^3b^-5
all to the 0 power.. would this just be 1?
I know it would simplify to
3b^10 but i dont know what to do with the 0 power
-----
a^8
Found 3 solutions by ccs2011, Alan3354, Theo:
Answer by ccs2011(207) About Me  (Show Source):
You can put this solution on YOUR website!
Yes any number other than 0 raised to zero power is 1
Also when you simplified, the 3 should be in denominator
4%2F12+=+1%2F3

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Anything to the power of 0 is 1, including zero (by convention).
By convention means we all agree to believe it.
-----------------
If you mean
%28%284a%5E-5b%5E5%29%2F%2812a%5E3b%5E-5+%29%29%5E0
= 1

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
it looks to me like your equation would be:

anything to the 0 power is equal to 1, so the whole thing should reduce to be equal to 1.
you can go through the exercise of simplifying the expression, but if the whole thing is raised to the 0 power, then the answer has to be 1.
i show the reduced formula to be equal to:
%28%28b%5E10%29%2F%283a%5E8%29%29%5E0
if you follow the rules of exponentiation through, this becomes:
b%5E%2810%2A0%29%2F%283%5E0%2Aa%5E%288%2A0%29%29 which becomes:
b%5E0%2F%283%5E0%2Aa%5E0%29 which becomes:
1%2F%281%2A1%29 which becomes:
1%2F1 which becomes:
1.