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Math Help - Help with rearranging formula

  1. #1
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    Help with rearranging formula

    Hello, Im looking for some guidance in rearranging formula. Need to ask if what I'm doing is right?

    1) 4 (^x+1) = 6 ^0.5x , find x
    ln4 (^x+1) = ln6(^0.5x)
    x+1(ln4) = 0.5xln6
    1.38x +1.38 = 0.895x
    1.38 = (0.895/1.38) x
    1.38 = 0.65x
    x= 0.65/1.38
    ???

    2) I = Imax (1-e ^-Rt/L) , R=30, L=0.01 - find t in seconds for current to reach 95%

    = I/Imax = 1- e -Rt/1
    =I/Imax+1 = -e -Rt/1
    -ln (I/Imax+1) = -Rt/1 lne
    -Rt/1= -ln(I/Imax+1)
    -Rt= -ln(I/Imax +1) (l)
    t= -ln (1/0.95 +1) (0.01)/-30
    ???

    3) PV^Y = K , make Y the subject
    V^Y = K/P
    lnV^Y = ln K/P
    YlnV = ln K/P
    Y = K/P / lnV

    4) V= Vmax exp -R/1 t , make t the subject
    V/Vmax = e -R/1 t
    ln V/Vmax = ln(e- R/1 t)
    ln V/Vmax = -R/1 t lne
    ln (V/Vmax) = -R/1 t
    t = ln (V/Vmax) L/ -R
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Tino1210 View Post
    Hello, Im looking for some guidance in rearranging formula. Need to ask if what I'm doing is right?

    1) 4 (^x+1) = 6 ^0.5x , find x
    ln4 (^x+1) = ln6(^0.5x)
    x+1(ln4) = 0.5xln6
    1.38x +1.38 = 0.895x
    1.38 = (0.895/1.38) x ....
    1.38 = 0.65x
    x= 0.65/1.38
    ???

    2) I = Imax (1-e ^-Rt/L) , R=30, L=0.01 - find t in seconds for current to reach 95%

    = I/Imax = 1- e^ -Rt/1
    =I/Imax+1 = -e^ -Rt/1
    -ln (I/Imax+1) = -Rt/1 lne
    -Rt/1= -ln(I/Imax+1)
    -Rt= -ln(I/Imax +1) (l)
    t= -ln (1/0.95 +1) (0.01)/-30
    ???

    3) PV^Y = K , make Y the subject
    V^Y = K/P
    lnV^Y = ln K/P
    YlnV = ln K/P
    Y = ln(K/P) / lnV just you forget blue ln

    4) V= Vmax exp -R/1 t , make t the subject
    V/Vmax = e -R/1 t
    ln V/Vmax = ln(e- R/1 t)
    ln V/Vmax = -R/1 t lne
    ln (V/Vmax) = -R/1 t
    t = ln (V/Vmax) L/ -R
    1)what you did here
     1.38x+1.38=0.895x

    <br />
\Rightarrow 1.38=.895x-1.38x\Rightarrow1.38=-.935x

    <br />
\Rightarrow x=\frac{-1.38}{.935}



    2) I/Imax = 1- e^ -Rt/1

    <br />
\Rightarrow \frac{I}{Imax}-1 = -e^{\frac{-Rt}{L}}<br />

    <br />
\Rightarrow 1-\frac{I}{Imax}=e^{\frac{-Rt}{L}}

    <br />
\Rightarrow ln(1-\frac{I}{Imax})= \frac{-Rt}{L}

    <br />
\Rightarrow \frac{(L)ln(1-\frac{I}{Imax})}{-R}=t


    the last one is not clear
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  3. #3
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    thanks for that, yeah the last one is hard to type out, sorry
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  4. #4
    MHF Contributor Amer's Avatar
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    what I can't understand is e to power what this is what you ask about

    V=(Vmax)e^{\frac{-R}{L}} t
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  5. #5
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    Quote Originally Posted by Amer View Post
    what I can't understand is e to power what this is what you ask about

    V=(Vmax)e^{\frac{-R}{L}} t
    yes, the t is positioned in the question between the -R/L
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  6. #6
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Amer View Post
    what I can't understand is e to power what this is what you ask about

    V=(Vmax)e^{\frac{-R}{L}} t

    V=(Vmax)e^{\frac{-R}{L}}t

    \frac{V}{Vmax}=e^{\frac{-R}{L}}t

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

    I think this what you are asking about right ?
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  7. #7
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    yeah thats it, thanks
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  8. #8
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    is this one correct:

    D = Do + log (x/y) , make x the subject

    D- Do = log (x/y)
    ln (D-Do) = x/y
    y ln (D-Do) = x
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  9. #9
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Tino1210 View Post
    is this one correct:

    D = Do + log (x/y) , make x the subject

    D- Do = log (x/y)
    ln (D-Do) = x/y
    y ln (D-Do) = x
    no it is wrong

    you know that

    log A = log_{10} (A) if you do not know that now you know it

    D-Do = log(x/y)

    10^{(D-Do)} = \frac{x}{y}

    y(10^{(D-Do)}) = x
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