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

1 Answer

Answer:

#x=1#

Explanation:

We can use some rules to find out.

First rule we'll use is #x^-1=1/x#:

#x^(-3/5)=1#

#1/(x^(3/5))=1/1#

We can invert both sides:

#x^(3/5)/1=1/1=>x^(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 #(-1)^2#, we end up with 1. Same with #(-1)^4, (-1)^6# and all other even exponents. Not so with #(-1)^3, (-1)^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}