# How do you solve x^(-3/5) = 1?

$x = 1$

#### Explanation:

We can use some rules to find out.

First rule we'll use is ${x}^{-} 1 = \frac{1}{x}$:

${x}^{- \frac{3}{5}} = 1$

$\frac{1}{{x}^{\frac{3}{5}}} = \frac{1}{1}$

We can invert both sides:

${x}^{\frac{3}{5}} / 1 = \frac{1}{1} \implies {x}^{\frac{3}{5}} = 1$

Now to the exponent. We're asked to cube $x$ (that's the 3) and also take the 5th root.

With odd exponents, sign is preserved. What I mean by that is that when we have ${\left(- 1\right)}^{2}$, we end up with 1. Same with ${\left(- 1\right)}^{4} , {\left(- 1\right)}^{6}$ and all other even exponents. Not so with ${\left(- 1\right)}^{3} , {\left(- 1\right)}^{5}$ and all other odd exponents - we end up with $- 1$.

What this means is that $x = 1$ but it can't equal $- 1$.

Here's a graph to show the solution:

graph{(y-x^-(3/5))((x-1)^2+(y-1)^2-.1)=0}