How to solve for x in this problem

  • Thread starter mmzaj
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In summary, the general solution for x in the given equation involves using Lambert's W function. The equation can be transformed into a form involving this function, but it may not be possible to find an exact-explicit solution. However, using the W function still provides a named expression for the solution rather than a purely numerical approximation.
  • #1
mmzaj
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what's the general solution for x in the following equation

[tex] x=c-N^{-\left(\frac{1+x}{1-x}\right)}[/tex]

i could solve using numerical methods , but i need an exact-explicit solution .
 
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  • #2
Looks like you'll need Lambert's W, then.
 
  • #3
You could substitute the brakets with x by a new variable and then try to rearrange until you can apply
http://mathworld.wolfram.com/LambertW-Function.html
In the end this new function is just like numeric. But at least you have a name for it ;)
 
  • #4
Gerenuk said:
You could substitute the brakets with x by a new variable and then try to rearrange until you can apply
http://mathworld.wolfram.com/LambertW-Function.html
In the end this new function is just like numeric. But at least you have a name for it ;)

thanks for the reply .. i tried this :
substitute

[tex]y=\frac{1+x}{1-x}[/tex]then

[tex]N^{-y}=c-\frac{y-1}{y+1}[/tex]

or

[tex](y+1)N^{-y}=(c-1)y+(c+1)[/tex]

but then i got a mental block ... i don't know how to transform the problem into the form :

[tex]A(N,c)=B(x) e^{B(x)} [/tex]
 
  • #5
i managed to reduce the problem down to the form :

[tex]z^{(z-A)}=B[/tex]

again , I'm stuck !
 
  • #6
Hmm, maybe it's not doable with that function alone. I only get
[tex]
W(x)=Ax+B
[/tex]
But anyway, what's the difference between numerics and knowing an expression with W? W has to be computed numerically anyway (just like sin, cos and all other functions).
 
  • #7
[tex]\pi[/tex] has to be computed numerically too but we still allow it a symbol for convenience. And I would much rather sin(2) over a numerical approximation for it, even though if anyone wanted to evaluate sin(2) they would end up with an approximation anyway.

I also couldn't find a solution in terms of the W function. I checked with Wolfram and it too couldn't find one.
 

FAQ: How to solve for x in this problem

How do I solve for x in a simple algebraic equation?

To solve for x in a simple algebraic equation, you need to isolate the variable on one side of the equation. You can do this by using the inverse operation of the given operation. For example, if the equation is 2x + 5 = 15, you would subtract 5 from both sides to get 2x = 10. Then, you would divide both sides by 2 to get x = 5.

How do I solve for x when there are variables on both sides of the equation?

In this case, you need to simplify the equation by combining like terms on each side. Then, you can use the same method as in the previous question to isolate the variable on one side of the equation. For example, if the equation is 2x + 5 = 3x + 10, you would subtract 2x from both sides to get 5 = x + 10. Then, you would subtract 10 from both sides to get -5 = x.

What should I do if there are fractions or decimals in the equation?

If there are fractions or decimals in the equation, you can first try to eliminate them by multiplying both sides of the equation by the common denominator. If that is not possible, you can convert the fractions into decimals or vice versa to make the equation easier to solve. For example, if the equation is 1/2x = 3, you can multiply both sides by 2 to get x = 6. If the equation is 0.5x = 3, you can multiply both sides by 10 to get x = 30.

What if there are parentheses in the equation?

If there are parentheses in the equation, you can use the distributive property to remove them. For example, if the equation is 2(x + 3) = 14, you would multiply 2 by both x and 3 to get 2x + 6 = 14. Then, you can solve for x using the methods mentioned in the previous questions.

Can I check my answer to make sure it is correct?

Yes, you can always check your answer by substituting it back into the original equation. If the equation holds true, then your answer is correct. If not, you may have made a mistake during the solving process and should try again.

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