SOLUTION: Can you tell me if I have simplified this equation correctly? (9y^2/(x^-8*y^0))^1/2 I first found that y^0 = 1. So that would drop out of the equation. Then I thought that

Algebra ->  Exponents -> SOLUTION: Can you tell me if I have simplified this equation correctly? (9y^2/(x^-8*y^0))^1/2 I first found that y^0 = 1. So that would drop out of the equation. Then I thought that      Log On


   

Question 43526: Can you tell me if I have simplified this equation correctly?
(9y^2/(x^-8*y^0))^1/2
I first found that y^0 = 1. So that would drop out of the equation.
Then I thought that raising to the 1/2 is the same as taking the square root of the equation.
So I now have the square root of 9y^2 / square root of x^-8.
The square root of 9y^2 is 3y. So I now have 3y/square root of x^-8.
I think that a negative exponent is the reciprocal of the positive so I now have 3y * square root of x^8. And x^8 is x^4 * x^4. So if I take the square root of x^4 is x^2. So I ended up with 3y*2x^2. Is that right?
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Let's see...you want to simplify:
%289y%5E2%2F%28x%5E%28-8%29%2Ay%5E0%29%29%5E%281%2F2%29
Your first step is correct:y%5E0+=+1, so now yu have:
%289y%5E2%2Fx%5E-8%29%5E%281%2F2%29
Your second statement is also correct:x%5E%281%2F2%29+=+sqrt%28x%29.
So now you want to multiply the internal exponents by the external exponent:
%289%5E%281%2F2%29%2A%28y%5E2%29%5E%281%2F2%29%29%2F%28x%5E-8%29%5E%281%2F2%29 Simplify this. For the numerator:9%5E%281%2F2%29+=+sqrt%289%29 = 3 and %28y%5E2%29%5E%281%2F2%29+=+sqrt%28y%5E2%29 = y
And for the denominator: %28x%5E%28-8%29%29%5E%281%2F2%29+=+x%5E%28-4%29
3y%2Fx%5E-4 Now bring the denominator up and change the sign of the exponent.
3x%5E4y