SOLUTION: help me solve this equation: {{{(2xy^-2y^4/3x^-1y)^-2 (4xy/2x^-1y^-3)^2}}}

Algebra ->  Exponents-negative-and-fractional -> SOLUTION: help me solve this equation: {{{(2xy^-2y^4/3x^-1y)^-2 (4xy/2x^-1y^-3)^2}}}      Log On


   



Question 26879: help me solve this equation: %282xy%5E-2y%5E4%2F3x%5E-1y%29%5E-2++++%284xy%2F2x%5E-1y%5E-3%29%5E2
Answer by bmauger(101) About Me  (Show Source):
You can put this solution on YOUR website!
This problem's not as bad as it looks, just work within each parentheses and remember all of your rules for exponents.
Taking a look at the inside of the first parentheses. Note that you can break it up into smaller parts by terms if the whole thing looks intimidating. So rather than think through:
%282xy%5E-2y%5E4%29%2F%283x%5E-1y%29 it can be seen as:
%282%2F3%29%2A%28%28x%5E1%29%2Fx%5E%28-1%29%29%2A%28y%5E%28-2%29y%5E4%2Fy%29
2/3 can't be reduced. For the rest of them you can always move anything on the bottom of the fraction to the top if you negate it's exponent.
So x%5E1%2Fx%5E%28-1%29 can be written x%5E1%2Ax%5E1=x%5E2.
The same thing can be done with your y's
y%5E-2y%5E4%2Fy=y%5E-2%2Ay%5E4%2Ay=y%5E%28-2%2B4%2B1%29=y%5E3
So your first parentheses reduces to:
%282x%5E2y%5E3%2F3%29%5E-2
When you raise this to a negative exponent like -2, it's usally easiest to just take the reciprical of the parentheses and then raise it to positive 2:
%283%2F2x%5E2y%5E3%29%5E2 Then simply take each term, and raise it to the second power to get:
9%2F4x%5E4y%5E6
Do the same thing to the second parentheses, then you'll have to combine/cancel terms between the two when you multiply. It's slightly time consuming, but not that hard & good practice. Again, just take your time, work the inside before trying to square anything, and keep your rules of exponents handy. Let me know if you have any further problems.