SOLUTION: I need some help with this problem: Simplify (3y^8/2zy^2)^4 Thanks on advance

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Question 170952: I need some help with this problem:
Simplify (3y^8/2zy^2)^4
Thanks on advance
Found 2 solutions by jim_thompson5910, angel1286:
Answer by jim_thompson5910(28704) About Me  (Show Source):
You can put this solution on YOUR website!
Note: %28x%5Ey%29%5Ez=x%5E%28y%2Az%29. Notice how the exponents are being multiplied.

Also, we can write 3 as 3%5E1, 2 as 2%5E1, and "z" as z%5E1 which will make the expression %28%283%5E1y%5E8%29%2F%282%5E1z%5E1y%5E2%29%29%5E4


%28%283%5E1y%5E8%29%2F%282%5E1z%5E1y%5E2%29%29%5E4 Start with the given expression


%283%5E1y%5E8%29%5E4%2F%282%5E1z%5E1y%5E2%29%5E4 Break up the fraction


%283%5E%281%2A4%29y%5E%288%2A4%29%29%2F%282%5E%281%2A4%29z%5E%281%2A4%29y%5E%282%2A4%29%29 Multiply the outer exponent by EVERY exponent inside the parenthesis


%283%5E%284%29y%5E%2832%29%29%2F%282%5E%284%29z%5E%284%29y%5E%288%29%29 Multiply the exponents.


%2881y%5E%2832%29%29%2F%2816z%5E%284%29y%5E%288%29%29 Raise 3 to the 4th power to get 81. Raise 2 to the 4th power to get 16.


%2881y%5E%2832-8%29%29%2F%2816z%5E%284%29%29 Now divide the "y" terms by subtracting their exponents


%2881y%5E%2824%29%29%2F%2816z%5E%284%29%29 Subtract


So %28%283y%5E8%29%2F%282zy%5E2%29%29%5E4 simplifies to %2881y%5E%2824%29%29%2F%2816z%5E%284%29%29

Answer by angel1286(3) About Me  (Show Source):
You can put this solution on YOUR website!
first simplify with the parenthesis: the 3 and the 2 cannot be divided so leave them like they are, but the "y" can be so according to power rules, in a division we would subtract the exponents so +y%5E8%2Fy%5E2+=+y%5E6
the the equation looks like +%283%2F2+%2A+y%5E6%29%5E4+=+%2881%2F16%29%2A+y%5E24 because the 3 and the 2 are raised to the fourth and the according to power rules, the power of a power multiplies each other.