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Math Help - Exponential Problem (Quite hard)

  1. #1
    Bar0n janvdl's Avatar
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    Exponential Problem (Quite hard)

    Here's a problem i really can't solve. I've tried logs and spent 2 science periods and an accounting period on it.

    Find values of  x for which the following equation would be true.

     x^2 = 2^x

    By inspection we can see x = 2 but i'm sure this isn't the only solution.

    Even my teacher can't solve this one. Am i missing something simple here?
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  2. #2
    MHF Contributor red_dog's Avatar
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    Sketching the graphs, there are three solutions:
    x_1\in(-1,0),x_2=2,x_3=4
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    Bar0n janvdl's Avatar
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    Quote Originally Posted by red_dog View Post
    Sketching the graphs, there are three solutions:
    x_1\in(-1,0),x_2=2,x_3=4
    Thank you Red_Dog, but i wish to see an algebraic solution please. I want to see how it should be solved.

    And please explain what you mean with this:  x_1\in(-1,0)
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    MHF Contributor red_dog's Avatar
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    The solutions of an equation of the form f(x)=g(x) are the x-coordinates of the points of intersection between the function's graphs.
    In the sketch, a point of intersection has the x-coordinate between -1 and 0. So, a solution of the equation is in that interval.
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  5. #5
    Bar0n janvdl's Avatar
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    Quote Originally Posted by red_dog View Post
    The solutions of an equation of the form f(x)=g(x) are the x-coordinates of the points of intersection between the function's graphs.
    In the sketch, a point of intersection has the x-coordinate between -1 and 0. So, a solution of the equation is in that interval.
    But i just cannot see a negative value satisfying the equation...
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  6. #6
    MHF Contributor red_dog's Avatar
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    Probably you can't find an exact negative solution, but this solution exists.
    Let f(x)=x^2-2^x
    We have f(-1)=1-\frac{1}{2}=\frac{1}{2}>0 and f(0)=-1<0.
    f is continue and it has the Darboux property. So f must have at least a solution in (-1,0).
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  7. #7
    Bar0n janvdl's Avatar
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    Quote Originally Posted by red_dog View Post
    Probably you can't find an exact negative solution, but this solution exists.
    Let f(x)=x^2-2^x
    We have f(-1)=1-\frac{1}{2}=\frac{1}{2}>0 and f(0)=-1<0.
    f is continue and it has the Darboux property. So f must have at least a solution in (-1,0).
    Is there no way of taking a pen and paper and working this out? All the solutions it has.
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  8. #8
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    Quote Originally Posted by janvdl View Post
    Here's a problem i really can't solve. I've tried logs and spent 2 science periods and an accounting period on it.

    Find values of  x for which the following equation would be true.

     x^2 = 2^x

    By inspection we can see x = 2 but i'm sure this isn't the only solution.

    Even my teacher can't solve this one. Am i missing something simple here?
    Yeah, sometimes you encounter such equations that can't be solved the "normal" way. I think there's a name for such equations.

    They can be solved graphically, as shown.

    And they can be solved analytically, or by "pen and paper" or calculator. It is by trial-and-error, by repetition, by iteration.

    Before I learned about the Newton's Method or Newton-Rhapson Method, I was already using iteration when I cannot solve quadratic/cubic/etc/trigo equations. . Newton's Method made it easier for me because it is systematic and after only about 3 iterations, depending on how close your seed root/factor is, you'd almost get the real root/factor.

    So,
    x^2 = 2^x
    x^2 -2^x = 0
    Let y = x^2 -2^x --------(i)
    Use Newton's Method of iteration on (i). If x would make y zero or very close to zero, then you got your x. There can be many of those x's, though.

    x2 = x1 -[y1 / (y1')] -------Newton's Method.
    where
    ---x1 is the seed x. The x you assume to be a root or factor.
    ---x2 is the next x that should be closer to the correct x than x1
    ---x3 is the next x that should be closer to the correct x than x2
    ---x4 is the next x that should be closer to the correct x than x3
    ...(iteration, you know. I usually stop at x3 or x4.)
    ---y1 is value of y when x = x1
    ---(y1)' is value of first derivative of y when x = x1.
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  9. #9
    Forum Admin topsquark's Avatar
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    I get x \approx -0.7666647.

    -Dan
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  10. #10
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    There are no general way to solve equations. The only thing we can show is that a solution exists.
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