SOLUTION: Here is the problem: (5/4)to the negative 3rd power (5/4) to the negative 2nd power (5/4)to the 2nd power I always thought that to get rid of a negative power, you invert. F

Algebra ->  Exponents-negative-and-fractional -> SOLUTION: Here is the problem: (5/4)to the negative 3rd power (5/4) to the negative 2nd power (5/4)to the 2nd power I always thought that to get rid of a negative power, you invert. F      Log On


   

Question 27137: Here is the problem:
(5/4)to the negative 3rd power (5/4) to the negative 2nd power (5/4)to the 2nd power
I always thought that to get rid of a negative power, you invert. For example the first one would be (4/5) to the 3rd power, the second one would be (4/5) to the 2nd power....but what do I do with the 3rd part? It is not a negative exponent. Then you would add/subtract the exponents and the base would remain (4/5). Please help if you understand my question.
Thanks,
Patty
Answer by askmemath(368) About Me  (Show Source):
You can put this solution on YOUR website!
There are 3 ways you can go about solving this Question.
The easiest way is already mentioned in your question itself and you havent even realised it!!
We have %285%2F4%29%5E%28-2%29 and %285%2F4%29%5E2
Since both have the same number i.e.5%2F4 all you had to do was add their exponents
But we know that -2 +2 = 0
So basically we are left with 1
%28%284%2F5%29%5E3%29%2A1 = 64%2F25

You've tried the 2nd way so let me finish that one too
Since you are writin everything in terms of 4/5,let's write the last term also(the one with the positive power)in terms of 4/5 as well
%284%2F5%29%5E%28-2%29
%28%284%2F5%29%5E3%29%28%284%2F5%29%5E2%29%284%2F5%29%5E%28-2%29
Adding the exponents only now we get 3-2+2 = 3
or %284%2F5%29%5E3 = 64%2F25
Same as above. Yay!!