Rearranging equation

  • Thread starter hexa
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I'm trying to rearange an equation but don't manage to get the right results so I guess the rearranging did not work out properly. Where is my mistake?

Z=Y+(X/W)Ve(^(WU) -1)
Y=0


my sollution:
U=ln (Z/((X/W)U))+1 /W

thanks a lot

hexa
 

VietDao29

Homework Helper
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hexa said:
Z=Y+(X/W)Ve(^(WU) -1)
I don't really get this equation. Can you write it in a clearer way? What's V, and is it e ^(WU - 1) or e ^(WU) - 1?
Viet Dao,
 

EnumaElish

Science Advisor
Homework Helper
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hexa said:
my sollution:
U=ln (Z/((X/W)U))+1 /W
What happens to V? Why is there a U on the right? Is that supposed to be V? And yeah, what does e(^(WU)-1) mean? Did you mean to write e^(WU-1)?
 

HallsofIvy

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Homework Helper
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hexa said:
I'm trying to rearange an equation but don't manage to get the right results so I guess the rearranging did not work out properly. Where is my mistake?

Z=Y+(X/W)Ve(^(WU) -1)
Y=0


my sollution:
U=ln (Z/((X/W)U))+1 /W

thanks a lot

hexa
Im not clear on what e(^(WU)-1) could mean. I'm going to assume that it was e^(WU-1) (eWU-1). The other possiblility is e^(WU)- 1 (eWU-1.

If Y= 0, then the equation is really Z= (X/W)V eWU-1. Divide both sides by (X/W)V and we have (WZ)/(XV)= eWU-1. Take the natural logarithm of both sides to get ln((WZ)/(XV))= WU-1. Add 1 to both sides: WU= 1+ ln((WZ)/(XV). Finally, divide both sides by W and you have
U= (1+ ln((WZ)/(XV)))/W.

Of the problem was, in fact, Z= (X/W)V (eWU-1) then start out the same as above- divide by (X/W)V to get (WZ)/(XV)= eWU-1.
Add 1 to both sides: eWU= 1+(WZ)/(XV). Take logarithms of both sides: WU= ln(1+ (WZ)/(XV)). Finally, divide both sides by W:
U= ln(1+(WZ)/(XV))/W.
 

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