# Rearrangement of two values

• Mar 9th 2013, 05:26 AM
bean11
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!
• Mar 9th 2013, 08:09 AM
MINOANMAN
Re: Rearrangement of two values
you do not need to be a mathematician to understand that what you are looking is .........IMPOSSIBLE.
• Mar 9th 2013, 10:25 AM
bean11
Re: Rearrangement of two values
Apologies, i didnt mean W*L, but simply both W and L as the subject!
• Mar 9th 2013, 11:14 AM
HallsofIvy
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.
• Mar 9th 2013, 11:27 AM
MINOANMAN
Re: Rearrangement of two values
OK if L or W separate it is possible as Hall show you above
• Mar 9th 2013, 01:07 PM
bean11
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?
• Mar 11th 2013, 01:50 AM
bean11
Re: Rearrangement of two values
Anyone?

Further help would be greatly appreciated. Thanks