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Math Help - Having trouble with this one.

  1. #1
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    Having trouble with this one.

    Guys,

    I am new here so please forgive my ignorance.

    Need help with the following:
    Y = A*ln(B*ec*x - D*Y)

    Where A, B, C, D are constants. I would like to determine the if there is a 20% change in "x" what is the corresponding change/impact on "Y"?

    Thanks.
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  2. #2
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    Re: Having trouble with this one.

    Quote Originally Posted by Joesph View Post
    Guys,

    I am new here so please forgive my ignorance.

    Need help with the following:
    Y = A*ln(B*ec*x - D*Y)

    Where A, B, C, D are constants. I would like to determine the if there is a 20% change in "x" what is the corresponding change/impact on "Y"?

    Thanks.
    Is it a 20% increase or decrease in x?
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  3. #3
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    Re: Having trouble with this one.

    A 20% increase in "x"
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  4. #4
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    Re: Having trouble with this one.

    If you increase x by 20%, you end up with x + 0.2x = 1.2x, so replace x with 1.2x and see what effect this has.
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  5. #5
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    Re: Having trouble with this one.

    Thanks for the quick replys, but that's the part I don't know how to solve.
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  6. #6
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    Re: Having trouble with this one.

    Show me what you've tried. Start by replacing x with 1.2x...
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  7. #7
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    Re: Having trouble with this one.

    I would need to solve for "Y" first. The problem is that I do not know how to do that since both the left and right sides contain "Y"
    Y = A*ln(B*ec*x - D*Y)
    if I simply put Y = A*ln(B*ec*1.2x - D*Y), I would still need a way to solve for "Y"
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  8. #8
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    Re: Having trouble with this one.

    Hello, Joesph!

    \text{I would need to solve for }y\text{ first.} . You can't!

    \text{We have: }\:y \;=\;A\ln(Be^{cx} - Dy)

    \text{We have a }transcendental\text{ equation.}
    y\text{ is both inside and outside of a log function.}

    \text{It cannot be solved for }y.
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