How can Lambert's equation be used to solve for x?

[psyclone]

Hi,

I had a little problem which I'd like to have answered by anyone who has a-little time to spend!

How to solve for x, given the following;

2Ln(x+4)+1=x ?

the solutions are x= -3.91.. & 5.50..

Wolfram-alpha solves the problem using 'Lamberts W-function', which makes sense because using the inverse property of logs with exponents will not simplify this problem at all.

But by using Using Lambert's Transcendental Equation, with the correct substitution of 'x' for another variable say 'u', I think it can be solved.

Ln u = a u^b {b = 1}

solution: u = exp [ -W(-a*b)/b ] , where W( ) is Lambert's W-function.

Your thoughts?

{http://mathworld.wol...W-Function.html }
{http://mathworld.wol...alEquation.html }

 

[HallsofIvy]

Since the given equation involves "ln" and Lambert's W function solves xe^x= constant, I would get the logarithm alone, then take the exponential of both sides. First simplify: 2ln(x+ 4)= x- 1 so ln(x+ 4)= (x- 1)/2. Then x+ 4= e^{(x- 1)/2}= e^{x/2}e^{-1/2} and write that as (x+ 4)e^{-x/2}= e^{-1/2}.

Now, let u= x+ 4 so that x= u- 4 and e^{-x/2}= e^{-(u- 4)/2}= e^{-u/2}e^{2}. In terms of u, the equation is now ue^{-u/2}e^{2}= e^{-1/2} or ue^{-u/2}= e^{-2- 1/2}= e^{-5/2}. Finally, let v= -u/2. Then u= -2v so we have -2ve^v= e^{-5/2} or ve^v= -\frac{e^{-3/2}}{2} and v= W\left(-\frac{e^{-3/2}}{2}\right)

 

[psyclone]

Thx for your post.
but..
W(-1/(2e^(-3/2))) should be W(-1/(2e^(-5/2))),
but your working is correct!