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Thread: transposition of formula

  1. #1
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    transposition of formula

    I've got the following formula

    X=k(1-e^(-t/T)

    i've managed as part of my assignment to find values for all of the constants or variables except t.

    so i now have the formula 10=20(1-e^(-t/0.1)

    could anybody show me how to transpose this formula for t ?
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  2. #2
    Behold, the power of SARDINES!
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    You bet...

    First we need to isolate the term containing the variable.

    $\displaystyle 10=20(1-e^{-t/.1})$

    $\displaystyle \frac{1}{2}=1-e^{-t/.1}$

    $\displaystyle -\frac{1}{2}=-e^{-t/.1}$

    $\displaystyle \frac{1}{2}=e^{-t/.1}$

    taking the natural log of both sides gives.

    $\displaystyle ln(1/2)=ln(e^{-t/.1})$

    $\displaystyle ln(1)-ln(2)= \frac{-t}{.1}ln(e) $


    $\displaystyle -ln(2)=\frac{-t}{.1}$

    solving for t gives...

    $\displaystyle .1ln(2)=t$
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  3. #3
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    Quote Originally Posted by al2308 View Post
    I've got the following formula

    X=k(1-e^(-t/T)

    i've managed as part of my assignment to find values for all of the constants or variables except t.

    so i now have the formula 10=20(1-e^(-t/0.1)

    could anybody show me how to transpose this formula for t ?
    $\displaystyle 10 = 20 \left ( 1 - e^{-t/0.1} \right )$

    $\displaystyle \frac{1}{2} = 1 - e^{-t/0.1}$

    $\displaystyle e^{-t/0.1} = \frac{1}{2}$

    $\displaystyle -\frac{t}{0.1} = ln \left ( \frac{1}{2} \right ) = - ln(2)$

    $\displaystyle t = 0.1 \cdot ln(2)$

    -Dan
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