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Math Help - Solving for t?

  1. #1
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    Solving for t?

    Hey all not sure if this should be in pre-calc or calc. I'm trying to solve the following equation (Von Bertalanffy) for T.

    L(t)=Li(1-e^(-k(T-To)))

    L(t)=Length at age t
    Li=Theoretical maximum length
    k=coefficient of growth
    To=Theoretical age at which growth =zero

    I have all my other parameters, I just need an equation where I can enter the numbers and find T. Unfortunately I was never great at math...

    Thanks for any help...
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  2. #2
    MHF Contributor Unknown008's Avatar
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    L_t=L_i (1-e^{-k(T-T_o)})

    Move the L's first;

    \frac{L_t}{L_i} = 1-e^{-k(T-T_o)}

    Move the 1 now.

    1-\frac{L_t}{L_i} = e^{-k(T-T_o)}

    Use natural log on both sides;

    ln(1-\frac{L_t}{L_i} )= ln(e^{-k(T-T_o)})

    ln(1-\frac{L_t}{L_i} )= -k(T-T_o)

    Divide by -k...

    \frac{ln(1-\frac{L_t}{L_i} )}{-k}= T-T_o

    Add To;

    \frac{ln(1-\frac{L_t}{L_i} )}{-k} + T_o= T

    T = \frac{ln(1-\frac{L_t}{L_i} )}{-k} + T_o

    Here you are
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  3. #3
    Newbie hungthinh92's Avatar
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    Problem  L(t)= L_i(1-e^{-k(T-T_0)}

    Answer:
     1-\frac {L(t)}{L_i} = e^{-k(T-T_0)}
     ln(1-\frac {L(t)}{L_i}) = ln(e^{-k(T-T_0)})
    Therefore,
     ln(1-\frac {L(t)}{L_i}) = -k(T-T_0)
     T= \frac{ln(1-\frac {L(t)}{L_i})}{-k} + T_0
    Last edited by hungthinh92; June 10th 2010 at 09:03 AM. Reason: :( Late...
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  4. #4
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    Yall are quick....Thanks!
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