How to solve for x in an equation

**florx** · April 26th 2010, 06:37 PM

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

**TKHunny** · April 26th 2010, 06:48 PM

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?

**florx** · April 26th 2010, 06:53 PM

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.

**harish21** · April 26th 2010, 07:12 PM

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.

**florx** · April 27th 2010, 10:43 PM

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

**Soroban** · April 28th 2010, 06:15 AM

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:

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

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

**TKHunny** · April 28th 2010, 08:29 AM

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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