I heard about the $W$ Lambert function yet what I can see on Wikipedia deserve to only resolve the equations of form $x imes e^x=lambda$. How deserve to I do?
Online solvers told me the solution was $x=e^-fracW(2)2$ but I don"t understand why...
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When attempting to resolve equations making use of the Lambert $W$ feature, one typically has to exploit the fact that different manipulations affect the "coefficient" in front of $e$ and also the exponent above $e$ differently.
In this case, you have actually $x^2$ in the exponent, so you desire that in the coreliable as well. We deserve to attempt squaring our equation and also check out where that gets us:$$x^2e^2x^2 = 1$$OK. This squared the coeffective yet doubled the exponent. So now we have actually $x^2$ both places. However, the exponent is $2x^2$ rather of $x^2$. So, we multiply our equaiton by $2$ to get$$2x^2e^2x^2 = 2$$Now we deserve to use the Lambert $W$ function to get$$2x^2 = W(2)\x = sqrtW(2)/2$$
Firstly, you desire to usage the fact that $x = e^ln x $, therefore the expression becomes$$x e^(e^lnx)^2 = 1, $$combining it over a single exponential$$Rightarrowhead e^lnx + e^2 ln x = 1.$$Taking the herbal logarithm, we then obtain $$ln x + e^2 ln x = 0 Rightarrowhead e^2 ln x = -ln x .$$Now we have the right to reararray the expression to gain it into a type usable by the Lambert W attributes by separating with by $exp(2 lnx)$ and multiplying by $2$$$Rightarrowhead 2 = -2 ln x e^-2 ln x$$$$ Rightarrow W(2) = -2 ln x$$$$Rightarrowhead x = e^-W(2)/2$$
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