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

1 Answer
bartholomew
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#.

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