# Is it possible to solve this equation using the Lambert function?

I am trying to solve the following equation, where $0 \leq a < b \leq 1$ are constants and $x \in (a,b)$; $$\frac{x-a}{b-a} = e^{-2\log(2)/(x+1)}$$ and stumbled across the Lambert W-function which I can use if I can transform my equation into something of the form $m = z^z$ where $m$ is some constant and $x$ some functions which can be expressed in terms of x. Is this possible?

I can get this far: $$\frac{x-a}{b-a} = e^{-2\log(2)/(x+1)} \Leftrightarrow [x+1 = w] \Leftrightarrow$$ $$\frac{w-a-1}{b-a} = e^{-2\log(2)/w} \Leftrightarrow [A = (b-a)e^{-2\log(2)}, B=-(a+1)] \Leftrightarrow$$ $$w+B = Ae^{1/w}$$ but do not know how to continue from this.

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I do not see how to solve even $w+1=e^{1/w}$ using the Lambert W function. – GEdgar Oct 25 '12 at 12:01
@GEdgar is in my opinion a knowledable person for your question. Just one remark. Are there some proofs for such type of equations that an algebraic solution can not be found? – Seyhmus Güngören Oct 25 '12 at 12:09
@malin I also checked MATLAB as well as Wolfram alpha Mathematica online version for your equation. Both couldnt output any algebraic solution. – Seyhmus Güngören Oct 25 '12 at 13:31

Your equation can be written as:

$$\frac{1}{y}+B=Ae^y\quad \mbox{with}\quad y=1/w$$ then

$$1+By=Aye^y\leftrightarrow \alpha y+ \beta=ye^y$$

call $z=\alpha y+\beta$ then we have

$$y=W(z)=W(\alpha y+\beta)$$

After this point I dont know $100\%$ if we can get $y$ alone or not. I had a similar problem as

$$y=W(e^x)+zx\quad$$

for which I was able to find $x=f(y)$. I suggest to check

Inverse function of $y=W(e^{ax+b})-W(e^{cx+d})+zx$

The difference is that in your case you have $W(y)$ and I had $W(e^y)$. If using the ideas over there, you can not get an algebraic solution, I am afraid that it is quite likely that there is no algebraic solution and one needs to solve it numerically. The numeric solution seems however simple as it defines the intersection of a line $ay+b$ with an $ye^y$. I hope this helps.

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