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Math Help - How to solve for x in an equation

  1. #1
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    Exclamation 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
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  2. #2
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    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?
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  3. #3
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    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.
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by florx View Post

    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.
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  5. #5
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    So how would we explain that point A is (1,1)? The back of the book says that is the answer.
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  6. #6
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    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


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  7. #7
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    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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