Rearranging Equation: Where is My Mistake?

  • Thread starter hexa
  • Start date
In summary, to rearrange the equation Z= Y+ (X/W)V (e^(WU)-1), if Y= 0, then the solution is U= (1+ ln((WZ)/(XV)))/W. If the equation is actually Z= (X/W)V (e^(WU)-1), then the solution is U= ln(1+ (WZ)/(XV))/W.
  • #1
hexa
34
0
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
 
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  • #2
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,
 
  • #3
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)?
 
  • #4
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

I am 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.
 

1. What is the purpose of rearranging an equation?

Rearranging an equation allows for a better understanding of the relationships between different variables and helps to solve for a specific variable.

2. Why is it important to check for mistakes when rearranging equations?

Mistakes in rearranging equations can lead to incorrect solutions and conclusions. It is important to check for mistakes to ensure the accuracy of the results.

3. How do I know if I made a mistake when rearranging an equation?

If the rearranged equation does not make logical sense or if the original equation cannot be derived from the rearranged equation, there is likely a mistake. Additionally, if the solution obtained from the rearranged equation does not match the expected result, there may be a mistake.

4. What are some common mistakes when rearranging equations?

Some common mistakes when rearranging equations include incorrect order of operations, forgetting to distribute negative signs, and making arithmetic errors.

5. How can I avoid making mistakes when rearranging equations?

To avoid making mistakes when rearranging equations, it is important to double check each step and use algebraic rules and properties correctly. It can also be helpful to simplify the equation before rearranging to make it easier to work with. Practice and familiarity with algebraic manipulations can also help to reduce chances of making mistakes.

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