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    I was reading about ordinary differential equations.

    The following lines appear in an equation in the book: ln=natural log

    -(L/R)ln(V-Ri) = t - (L/R)lnV

    (L/R)ln(V/(V-Ri)) = t

    V/(V-Ri) = e (to the power of Rt/L)


    Can anyone explain the steps taken going from line 1 to 2 and 2 to 3.

    If there's any missing lines, could you let me know. I can't seem to figure out how they've taken these steps.

    Thank you!
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    (Original post by little pixie)
    I was reading about ordinary differential equations.

    The following lines appear in an equation in the book: ln=natural log

    -(L/R)ln(V-Ri) = t - (L/R)lnV

    (L/R)ln(V/(V-Ri)) = t

    V/(V-Ri) = e (to the power of Rt/L)


    Can anyone explain the steps taken going from line 1 to 2 and 2 to 3.

    If there's any missing lines, could you let me know. I can't seem to figure out how they've taken these steps.

    Thank you!
     \dfrac{L}{R}\ln(\frac{V}{V-R_i}) = t

    Multiply both sides by R, divide by L:

     \ln(\frac{V}{V-R_i}) = \dfrac{Rt}{L}

    Take exponentials of both sides

     \dfrac{V}{V-R_i} = e^{\frac{Rt}{L}}

    EDIT

    and for step 1-2:

     \dfrac{-L}{R}\ln({V-R_i}) = t - \dfrac{L}{R}\ln(V)

    so

     \dfrac{-L}{R}\ln({V-R_i}) + \dfrac{L}{R}\ln(V) = t

    then factorise L/R

     \dfrac{L}{R}[\ln(V) - \ln({V-R_i})] = t

    use that  \ln(a) - \ln(b) = \ln(\frac{a}{b}) to get

     \dfrac{L}{R}[\ln(\frac{V}{V-R_i})] = t
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    From step 1 to 2. The log term is being brought from the right hand side to the left and a common factor of (L/R) is being take out of both log terms. Then by using the laws of logs: ln A - ln B = ln (A/B).

    From step 2 to 3 (L/R) is being brought to the right hand side of the equation and then both sides are being raised as powers of e.

    I hope you can follow this
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    (Original post by little pixie)
    I was reading about ordinary differential equations.

    The following lines appear in an equation in the book: ln=natural log

    -(L/R)ln(V-Ri) = t - (L/R)lnV

    (L/R)ln(V/(V-Ri)) = t

    V/(V-Ri) = e (to the power of Rt/L)


    Can anyone explain the steps taken going from line 1 to 2 and 2 to 3.

    If there's any missing lines, could you let me know. I can't seem to figure out how they've taken these steps.

    Thank you!
    From 1 to 2, they have added \frac{L}{R}\ln V to both sides and recalled the law of logs which states that a\ln (b) - a\ln c = a\ln (\frac{b}{c}) for use on the new LHS following the addition I mentioned.

    From 2 to 3, they rearrange first for \ln \frac{V}{V-Ri} and then note that e^{\ln x} = x so they raised both sides as a power of e to get rid of the log.
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    Thanks very much guys. That's a great help. It looks a lot easier now once explained!
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    Re write
    -(\frac{L}{R})ln(V-Ri)=t-(\frac{L}{R})lnV

    as
    -(\frac{L}{R})ln(V-Ri)+(\frac{L}{R})lnV=t

    Factor out the -(L/R) on the left side to get

    -(\frac{L}{R})(ln(V-Ri)-lnV)=t

    when you take the log of one thing away from the log of another thing it's the same as taking the log of their quotient. so....

    -(\frac{L}{R})ln(\frac{V-Ri}{V})=t

    getting closer now, lets add -(L/R) to both sides of the equation and raise e to the power of both sides of the equation the get rid of the natural logs.

    ln(\frac{V-Ri}{V})=t+\frac{L}{R}
    \frac{V-Ri}{V}=e^{t+\frac{L}{R}}


    Hope this helps you understand....Damn I hate LaTeX.
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    (Original post by little pixie)
    I was reading about ordinary differential equations.

    The following lines appear in an equation in the book: ln=natural log

    -(L/R)ln(V-Ri) = t - (L/R)lnV

    (L/R)ln(V/(V-Ri)) = t

    V/(V-Ri) = e (to the power of Rt/L)


    Can anyone explain the steps taken going from line 1 to 2 and 2 to 3.

    If there's any missing lines, could you let me know. I can't seem to figure out how they've taken these steps.

    Thank you!
    [edit]sorry double post[/edit]
 
 
 
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