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I wanted to solve the following equation.$$x \times e^{x^2}=1$$

I heard about the $W$ Lambert function but what I can see on Wikipedia can only solve the equations of type $x \times e^{x}=\lambda$. How can I do?

Online solvers told me the solution was $x=e^{-\frac{W(2)}{2}}$ but I don't understand why...

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2 Answers

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When attempting to solve equations using the Lambert $W$ function, one usually has to exploit the fact that different manipulations affect the "coefficient" in front of $e$ and the exponent above $e$ differently.

In this case, you have $x^2$ in the exponent, so you want that in the coefficient as well. We can try squaring our equation and see where that gets us:$$ x^2e^{2x^2} = 1 $$OK. This squared the coefficient but doubled the exponent. So now we have $x^2$ both places. However, the exponent is $2x^2$ instead of $x^2$. So, we multiply our equaiton by $2$ to get$$ 2x^2e^{2x^2} = 2 $$Now we can apply the Lambert $W$ function to get$$ 2x^2 = W(2)\\ x = \sqrt{W(2)/2} $$

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Firstly, you want to use the fact that $x = e^{ln x }$, therefore the expression becomes$$x e^{(e^{lnx})^2} = 1, $$combining it over a single exponential$$\Rightarrow e^{lnx + e^{2 ln x}} = 1.$$Taking the natural logarithm, we then get $$ln x + e^{2 ln x} = 0 \Rightarrow e^{2 ln x} = -ln x .$$Now we can rearrange the expression to get it into a form usable by the Lambert W functions by dividing through by $exp(2 lnx)$ and multiplying by $2$$$\Rightarrow 2 = -2 ln x e^{-2 ln x}$$$$ \Rightarrow W(2) = -2 ln x$$$$\Rightarrow x = e^{-W(2)/2}$$

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