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Math Help - Rearranging exponential equation

  1. #1
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    Rearranging exponential equation

    Hi,

    I've tried this on a Maths teacher friend of mine, who is currently stumped, so I hope it classes as 'advanced'. Essentially, I'm trying to rearrange the following equation:

    x = y / [A - e^(By)]

    to express y in terms of x. Is this possible?

    Thanks.
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  2. #2
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    Re: Rearranging exponential equation

    Hey mpylon.

    What I will do is use implicit differentiation to see if you can separate the variables so that they can be calculated independent of each other.

    d/dx(Ax - xe^(By)) = d/dx(y)

    dy/dx
    = A - d/dx(xe^(By))
    = A - [e^(By) + Bx*e^(By)dy/dx] So this implies:

    dy/dx[1 + Bxe^(By)] = A - e^(By) or

    dy/dx = [A - e^(By)]/[1 + Bxe^(By)]

    This looks like all derivatives will involve x's and y's which means that you can't get y as a function purely of x (or the other way around unless you use specific restrictions or find a parameterization.
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  3. #3
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    Re: Rearranging exponential equation

    Quote Originally Posted by chiro View Post
    Hey mpylon.

    What I will do is use implicit differentiation to see if you can separate the variables so that they can be calculated independent of each other.

    d/dx(Ax - xe^(By)) = d/dx(y)

    dy/dx
    = A - d/dx(xe^(By))
    = A - [e^(By) + Bx*e^(By)dy/dx] So this implies:

    dy/dx[1 + Bxe^(By)] = A - e^(By) or

    dy/dx = [A - e^(By)]/[1 + Bxe^(By)]

    This looks like all derivatives will involve x's and y's which means that you can't get y as a function purely of x (or the other way around unless you use specific restrictions or find a parameterization.
    Why are you differentiating at all? It should be obvious that y can't be isolated because it appears as both an exponent and a polynomial...
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  4. #4
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    Re: Rearranging exponential equation

    The point is to show that the functions change depends on its current value which shows its implicit.

    Recall that a function is uniquely determined by its Taylor series and one way to show implicit behavior is through its derivative.
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  5. #5
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    Re: Rearranging exponential equation

    There's a similar problem here.

    Solution of equation of form ln(a+bx) = cx
    Thanks from mpylon
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  6. #6
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    Re: Rearranging exponential equation

    Thanks "a tutor", the Lambert's W function was the way to go, although below a certain number the Lambert's W function gives complex numbers. Is there any way round this?
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  7. #7
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    Re: Rearranging exponential equation

    No worries, it was a quirk of the Excel function I downloaded, but have solved it. Thanks for your help.
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