I need to solve a function for $x$
$k+\frac{1}{2}=-\log{x}+\frac{x^2}{2}$
I know the solution(s) should be expressed using Lambert W function because the above function can also be expressed as
$2^{k+1/2}=\frac{1}{x}e^{(\ln{\sqrt{2}})x^2}$
however, I cannot proceed any further.
By log rules, we have
$$2^{-\log_2(x)+\frac{x^2}2}=\frac{2^{\frac{x^2}2}}{2^{\log_2(x)}}=\frac1x2^{\frac{x^2}2}=2^{k+\frac12}$$
Multiply both sides by $x$ and square both sides:
$$2^{x^2}=x^22^{2k+1}$$
Let $u=x^2$ to simplify:
$$2^u=u2^{2k+1}$$
Divide both sides by $2^{u+2k+1}$ and multiply both sides by $-\ln(2)$.
$$-\ln(2)2^{-2k-1}=-\ln(2)u2^{-u}=-\ln(2)ue^{-\ln(2)u}$$
Take the $W$ of both sides to get
$$W(-\ln(2)2^{-2k-1})=-\ln(2)u$$
$$u=\frac{W(-\ln(2)2^{-2k-1})}{-\ln(2)}$$
So finally,
$$x=+\sqrt{\frac{W(-\ln(2)2^{-2k-1})}{-\ln(2)}}$$
