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Algebraic indices

A quick question, how do you solve:

(2u4/5v3)-1?
Original post by Fizzle-A
A quick question, how do you solve:

(2u4/5v3)-1?


You can't because it's not equal to anything and you're not specifying which variable to solve for.

If you mean to simplify it, then just take the reciprocal of the fraction.
Reply 2
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Fizzle-A
OP
Original post by RDKGames
You can't because it's not equal to anything and you're not specifying which variable to solve for.

If you mean to simplify it, then just take the reciprocal of the fraction.


Yes, I meant simplify.

(2u4/5v3)-1
= (5v3/2u4)1

???
Reply 3
Avatar for Fizzle-A
Fizzle-A
OP
How do I simplify:

(32x10y-5)-1/5

(5√1/32) x (5√1/x10) x (5√1/y5)
??

Is this correct so far? I don't think it is as the power is -1/5 ...
(edited 7 years ago)
Reply 4
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B_9710
18
Original post by Fizzle-A
How do I simplify:

(32x10y-5)-1/5

(5√1/32) x (5√1/x10) x (5√1/y5)
??

Is this correct so far? I don't think it is as the power is -1/5 ...


(32x10y−5)−1/5≡32−1/5⋅(x10)−1/5⋅(y−5)−1/5 (32x^{10}y^{-5})^{-1/5}\equiv 32^{-1/5}\cdot (x^{10})^{-1/5}\cdot (y^{-5})^{-1/5} . Now just simplify each separately and you have the answer.

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