# Thread:

1. ## How to solve for x in an equation

Solve for x.
x^n = x^(1/n)

How would we solve for x in this case? Would we move the x^(1/n) over to the left side and set it to zero. Then we factor out the left side? Please help me with this. Thanks

2. Why have you encountered this equation? Perhaps it is only a thought / property question. Rephrase in words and think about it.

What number is alterered identically by a power and a root.

The square is the square root.
The cube is the cube root.

There are not too many candidates.

Does it make a difference if x is positive or negative?
Does it make a difference if n is odd or even?

Again, why have you encountered this equation?

3. This is the problem. I am stuck on b). I have concluded that in order to find point A we must set x^n = x^(1/n) equal to each other because that is where they intersect. After finding x we can plug it back in the equation to find y and thus we would have a coordinate. But the problem is I don't know how to find x. Is there a better way to solve this or am I on the right track? Thanks for the help.

4. Originally Posted by florx

This is the problem. I am stuck on b). I have concluded that in order to find point A we must set x^n = x^(1/n) equal to each other because that is where they intersect. After finding x we can plug it back in the equation to find y and thus we would have a coordinate. But the problem is I don't know how to find x. Is there a better way to solve this or am I on the right track? Thanks for the help.
In the given graphs, suppose n =2. So, In the parabola $y = x^2$, the coordinate pairs are $(x, x^2)$. You should be able to see that the following points are on the graph: (1, 1), (−1, 1), (2, 4), (−2, 4), and so on.

The graph of the square root function is related to $y = x^2$. The coördinate pairs are $(x,\sqrt{x})$. For example, (1, 1), (4, 2), (9, 3), and so on.

So for any n, the graph of $y=x^{\frac{1}{n}}$ will not be negative.

5. So how would we explain that point A is (1,1)? The back of the book says that is the answer.

6. Hello, florx!

Your game plan is absolutely correct!

Solve for $x\!:\;\;x^n \:=\: x^{\frac{1}{n}}$

We have: . $x^n - x^{\frac{1}{n}} \;=\;0$

Factor: . $x^{\frac{1}{n}}\left(x^{\frac{n^2-1}{n}} - 1\right) \;=\;0$

And we have two equations to solve:

. . $x^{\frac{1}{n}} \:=\:0 \quad\Rightarrow\quad x \:=\:0^n \quad\Rightarrow\quad \boxed{x \:=\:0}$

. . $x^{\frac{n^1-1}{n}} - 1 \:=\:0 \quad\Rightarrow\quad x^{\frac{n^2-1}{n}} \:=\:1 \quad\Rightarrow\quad x \:=\:1^{\frac{n}{n^2-1}} \quad\Rightarrow\quad\boxed{x \:=\:1}\;\;\text{ for }n \neq \pm1$

7. That's just a little insane.

Back to my original question:
What number is alterered identically by a power and a root?

There are not too many candidates.
Only 0 and 1. finis.

Why drag through messy algebra with ambiguous exponents when it's really just a definition / property question?
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