# Thread: how to solve for x in this problem

1. ## how to solve for x in this problem

what's the general solution for x in the following equation

$\displaystyle x=c-N^{-\left(\frac{1+x}{1-x}\right)}$

i know that the lambert W function should be involved , but i don't know how to transform the problem in hand to the form

$\displaystyle A(N,c)=B(x) e^{B(x)}$

2. First thing I'd do is subtract x from both sides, and add the exponential to both sides. What do you get from doing that?

3. Originally Posted by Ackbeet
First thing I'd do is subtract x from both sides, and add the exponential to both sides. What do you get from doing that?
i'm not following you ... the problem will still the same , won't it !?

i tried this : substitute $\displaystyle y=\frac{1+x}{1-x}$

then :
$\displaystyle N^{-y}=c-\frac{y-1}{y+1}$
or:

$\displaystyle (y+1)N^{-y}=(c-1)y+(c+1)$

and then i get a mental block

4. I can transform your equation with the following changes:

$\displaystyle \displaystyle{x=c-N^{-\frac{1+x}{1-x}}}$

$\displaystyle \displaystyle{N^{-\frac{1+x}{1-x}}=c-x}$

$\displaystyle \displaystyle{-\frac{1+x}{1-x}\,\ln(N)=\ln(c-x)}$

$\displaystyle \displaystyle{-(1+x)\ln(N)=(1-x)\ln(c-x)}$

$\displaystyle \displaystyle{-\ln(N)-x\ln(N)=\ln(c-x)-x\ln(c-x)}$

$\displaystyle \displaystyle{-\ln(N)=\ln(c-x)-x\ln(c-x)+x\ln(N)}$

$\displaystyle \displaystyle{\ln(N)=x\ln(c-x)-x\ln(N)-\ln(c-x)}$

$\displaystyle \displaystyle{N=\frac{(c-x)^{x}}{N^{x}(c-x)}}$

$\displaystyle \displaystyle{N=\frac{(c-x)^{x-1}}{N^{x-1+1}}}$

$\displaystyle \displaystyle{N=\frac{(c-x)^{x-1}}{N\cdot N^{x-1}}}$

$\displaystyle \displaystyle{N^{2}=\left(\frac{c-x}{N}\right)^{\!\!x-1}}.$

This may or may not be any closer to the product log function that you're after. But I have managed to reduce the number of appearances of the variable x. Does this help?

One avenue that might be fruitful at this point is to write the equation as the following:

$\displaystyle \displaystyle{N^{2}=e^{(x-1)\ln\left(\frac{c-x}{N}\right)}}.$

These are just ideas. Take them or leave them. The solution to your problem, which I don't see at the moment, is not obvious, at least in my opinion.

5. ok ... i managed to reduce the problem down to the form :

$\displaystyle z^{(z-A)}=B$

again , i'm stuck

6. Hmm. There's no known closed-form solution to this equation that I'm aware of. Are you sure the product log function is involved here?

Why can't you resort to numerical methods at this point? Your function would respond well to Newton's method, I think, at least for certain values of A and B.