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

May 23, 2018

All $x \ne 0 \in \mathbb{R}$.

#### Explanation:

We have:

$- \frac{1}{x} ^ 5 = \frac{1}{{\left(- x\right)}^{5}}$.

Observe that for every value of $x \ne 0$ 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 $x < 0$,

$- \frac{1}{x} ^ 5 = \frac{1}{{\left(- x\right)}^{5}} \Rightarrow - \frac{1}{- x} ^ 5 = \frac{1}{{\left(- \left(- x\right)\right)}^{5}}$,

and from what we previously observed,

$- \frac{1}{- x} ^ 5 = \frac{1}{{\left(- \left(- x\right)\right)}^{5}} \Rightarrow \frac{1}{x} ^ 5 = \frac{1}{x} ^ 5$.

The same is true if $x > 0$,

$- \frac{1}{x} ^ 5 = \frac{1}{{\left(- x\right)}^{5}} \Rightarrow - \frac{1}{x} ^ 5 = - \frac{1}{x} ^ 5$.

Therefore this equality is true for all $x \ne 0 \in \mathbb{R}$.