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Math Help - Rearrangement of two values

  1. #1
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    Rearrangement of two values

    Hello,

    I am far from being a mathematician and as such am struggling rearranging the following equation as such as would make W*L the subject:

    I =2*n*F*C*D*L*[1/ln(4*D*t /W^2)]

    I'd be very grateful for the rearrangement of this and more importantly an explanation as to the best method to do so!

    Thank you very much in advance!
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  2. #2
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    Re: Rearrangement of two values

    you do not need to be a mathematician to understand that what you are looking is .........IMPOSSIBLE.
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  3. #3
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    Re: Rearrangement of two values

    Apologies, i didnt mean W*L, but simply both W and L as the subject!
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  4. #4
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    Re: Rearrangement of two values

    Well, you can't have "both" W and L as subjects but you can have either W or L.

    To solve for L, from I =2*n*F*C*D*L*[1/ln(4*D*t /W^2)] multiply both sides by that denominator, ln(4Dt/W^2), to get
    Iln(4Dt/W^2)= 2nFCDL
    and then ust divide by 2nFCD to leave on "L" on the right:
    2InFCDL ln(4Dt/W^2)= L

    To solve for W, start, as before, by multiplying both sides by ln(4Dt/W^2):
    Iln(4Dt/W^2)= 2nFCDL

    Now use the properties of the logarithm on the left to get I(ln(4Dt)- 2ln(W))= Iln(4Dt)- 2Iln(W)= 2nFCDL
    Subtract Iln(4Dt) from both sides: -2Iln(W)= 2nFCDL- Iln(4Dt)

    Divide both sides by -2I: ln(W)= -nFCDL/I- ln(4Dt) and, finally, take the exponential of both sides

    W= e^{-FCDL/I- ln(4Dt)}= (e^{-FCDL/I})/e^{ln(4Dt)}= e^{-FCDL/i}/4Dt.
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  5. #5
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    Re: Rearrangement of two values

    OK if L or W separate it is possible as Hall show you above
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  6. #6
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    Re: Rearrangement of two values

    ok that makes sense, to an extent with my limited knowledge. Thanks very much! My only further query is as to how the above is possible is say W and L are both unknown values, with the remaining values being known?
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  7. #7
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    Re: Rearrangement of two values

    Anyone?

    Further help would be greatly appreciated. Thanks
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