For which non-zero real values of x is -x^-5 = (-x)^-5 ?

1 Answer
May 23, 2018

All x!=0 in RR.

Explanation:

We have:

-1/(x)^5=1/((-x)^5).

Observe that for every value of x!=0 in x^5, if x is negative, then x^5 is negative; the same is true if x is positive: x^5 will be positive.

Therefore we know that in our equality, if x<0,

-1/(x)^5=1/((-x)^5) rArr -1/(-x)^5=1/((-(-x))^5),

and from what we previously observed,

-1/(-x)^5=1/((-(-x))^5) rArr 1/x^5=1/x^5.

The same is true if x>0,

-1/(x)^5=1/((-x)^5) rArr -1/x^5=-1/x^5.

Therefore this equality is true for all x!=0 in RR.