Hi,

I had a little problem which I'd like to have answered!

 

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.wolfram.com/LambertW-Function.html }

{http://mathworld.wolfram.com/LambertsTranscendentalEquation.html }

 

 

Well if I just wanted the roots in a hurry, I would probably go for a numeric method.

 

I haven't fully worked out an analytical solution but some thoughts are

 

You want to get rid of the (x+4) so look for a trig substitution

 

[latex](x+4) = \frac{2}{\pi}(\frac{\pi}{2}x + 2\pi)[/latex]

 

substitute [latex]y = \frac{\pi}{2}x[/latex]

 

[latex]=\frac{2}{\pi}(y + 2\pi)[/latex]

 

and note that

 

[latex]sin\beta(y + 2\pi) = sin(\beta y)[/latex]

 

 

Interesting way to simplify the problem,

 

but the domain for inverse of sin(nx); is [-1 to 1].

Solved it!

 

[latex]2 ln{(x+4)}= x-1[/latex]

 

Substitute, u = x + 4

 

[latex]\ln u = \frac{u - 5}{2}[/latex]

 

[latex]u = e^{\frac{u-5}{2}}[/latex]

 

[latex]ue^{-\frac{u}{2}}=e^{-\frac{5}{2}}[/latex]

 

[latex]-\frac{u}{2}e^{-\frac{u}{2}}=-\frac{e^{-\frac{5}{2}}}{2}[/latex]

 

[latex]Y=Xe^{X} \Rightarrow X=W(Y)[/latex] (Lambert's W function)

 

[latex]W(-\frac{1}{2e^{\frac{5}{2}}})=-\frac{u}{2}[/latex]

 

 

[latex]W_{0}(-\frac{1}{2e^{\frac{5}{2}}})=-0.043[/latex]

[latex]W_{-1}(-\frac{1}{2e^{\frac{5}{2}}})=-4.752[/latex]

 

Substitute x for u

 

[latex]x=-2W(-\frac{1}{2e^{\frac{5}{2}}})-4[/latex]

 

x = -3.914, 5.504