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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