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Thread: Derivatives: Solving for t -Xe^(-xt) + Ye^(-yt)

  1. #1

    Derivatives: Solving for t -Xe^(-xt) + Ye^(-yt)

    I have this equation and I know what the final answer should be but I am stuck on how to get to it using derivatives. Thanks in advance
    Solving for t

    -Xe-xt + Ye-yt


    Final answer should be
    t= ln(y/x)(1/(y-x))

  2. #2
    Elite Member
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    Quote Originally Posted by mrooney View Post
    I have this equation and I know what the final answer should be but I am stuck on how to get to it using derivatives. Thanks in advance
    Solving for t

    -Xe-xt + Ye-yt


    Final answer should be
    t= ln(y/x)(1/(y-x))
    That is not an equation - there is no "equal to"(=) sign in there.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #3
    Quote Originally Posted by Subhotosh Khan View Post
    That is not an equation - there is no "equal to"(=) sign in there.
    sorry, its supposed to be equal to 0

  4. #4
    Elite Member
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    Quote Originally Posted by mrooney View Post
    sorry, its supposed to be equal to 0
    So edit your original post.

    Also, please explain how are X, x, Y & y are related.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  5. #5
    Quote Originally Posted by mrooney View Post
    I have this equation and I know what the final answer should be but I am stuck on how to get to it using derivatives. Thanks in advance
    Solving for t

    -Xe-xt + Ye-yt


    Final answer should be
    t= ln(y/x)(1/(y-x))
    Assuming x and X are meant to be the same thing (and that's a BIG assumption) and the same for the Y and y:

    [tex]-xe^{-xt} + ye^{-yt} =0 [/tex]

    [tex] y*e^{-yt} = x* e^{-xt} [/tex]

    [tex]\frac {y}{x} = \frac {e^{-xt}}{e^{-yt}} [/tex]

    [tex]\frac{y}{x} = e^{yt-xt}[/tex]

    Now take natural log of both sides and see if you can make t the subject.
    Heavens to Murgatroyd!!

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