SOLUTION: How do you solve: (x^10*y^(-8))^-3 that is x to the exponent of 10; y to the exponent of -8; and all of it to the -3

Algebra ->  Exponents -> SOLUTION: How do you solve: (x^10*y^(-8))^-3 that is x to the exponent of 10; y to the exponent of -8; and all of it to the -3      Log On


   

Question 255642: How do you solve:
(x^10*y^(-8))^-3
that is x to the exponent of 10; y to the exponent of -8; and all of it to the -3
Found 3 solutions by stanbon, Alan3354, richwmiller:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
You can't.
You don't have an equation.
And you have two variables.
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Cheers,
Stan H.

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
How do you solve:
(x^10y^-8)^-3
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It's not an equation, so it can't be solved, only simplified.
= x%5E-30%2Ay%5E24
= y%5E24%2Fx%5E30
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PS use ^ for exponents, above the 6.

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
Very interesting problem. Easy solution.
%28x%5E10%2Ay%5E%28-8%29%29%5E%28-3%29
The important thing to remember is negative exponent means the reciprocal.
3%5E%28-3%29 = 1%2F%283%5E3%29

1%2F3%5E%28-3%29 = 1%2F%281%2F%283%5E3%29%29=3%5E3
We have reciprocals of reciprocals in your problem.
%28x%5E2%29%5E5=x%5E10

so
%28x%5E10%2Ay%5E%28-8%29%29%5E%28-3%29=
1%2F%28x%5E10%2Ay%5E%28-8%29%29%5E3=
1%2F%28x%5E10%2Fy%5E%288%29%29%5E3=
Let's deal with the reciprocal now
%28y%5E%288%29%2Fx%5E10%29%5E3=
%28y%5E%2824%29%2Fx%5E30%29