We can use a procedure known as "bootstrapping" to determine an approximation for the Lambert $W$ function. Let's go back to its definition.
For $x > 0$ the equation
$$
we^w = x
$$
has exactly one positive solution $w = W(x)$ which increases with $x$. Note that $(w,x) = (1,e)$ is one such solution, so if $x > e$ then $w > 1$. By taking logarithms of both sides of the equation we get
$$
\log w + w = \log x
$$
or
$$
w = \log x - \log w. \tag{1}
$$
When $x > e$ we therefore have
$$
w = \log x - \log w < \log x.
$$
In other words, our first approximation is that
$$
1 < w < \log x \tag{2}
$$
when $x > e$. We then have
$$
0 < \log w < \log\log x,
$$
and plugging this into $(1)$ yields
$$
\log x - \log \log x < w < \log x, \tag{3}
$$
where the left side is positive for $x > 1$. Taking logarithms as before yields
$$
\log\log x + \log\left(1 - \frac{\log\log x}{\log x}\right) < \log w < \log\log x,
$$
and upon substituting this back into $(1)$ we get
$$
\log x - \log\log x < w < \log x - \log\log x - \log\left(1 - \frac{\log\log x}{\log x}\right).
$$
Since $w = W(x)$ we have shown that
$$
\log x - \log\log x < W(x) < \log x - \log\log x - \log\left(1 - \frac{\log\log x}{\log x}\right) \tag{4}
$$
for $x > e$.
In your particular case we're interested in $W(e^{x+a})$, for which we have
$$
x+a - \log(x+a) < W(e^{x+a}) < x+a - \log(x+a) - \log\left(1 - \frac{\log(x+a)}{x+a}\right)
$$
for $x+a > 1$. In this sense we have
$$
W(e^{x+a}) \approx x+a - \log(x+a) = x\left(1 - \frac{\log(x+a) - a}{x}\right) \tag{5}
$$
when $x+a$ is large. Now by applying Taylor series a couple times we see that, for $x$ large and $a \ll x$,
$$
\begin{align}
\frac{\log x - a}{x+1} &= \frac{\log x - a}{x} \cdot \frac{1}{1+\frac{1}{x}} \\
&\approx \frac{\log x - a}{x} \left(1-\frac{1}{x}\right) \\
&= \frac{\log x - a}{x} - \frac{\log x - a}{x^2} \\
&= \frac{\log(x+a-a) - a}{x} - \frac{\log x - a}{x^2} \\
&= \frac{\log(x+a) + \log\left(1-\frac{a}{x+a}\right) - a}{x} - \frac{\log x - a}{x^2} \\
&= \frac{\log(x+a) - a}{x} + \frac{\log\left(1-\frac{a}{x+a}\right)}{x} - \frac{\log x - a}{x^2} \\
&\approx \frac{\log(x+a) - a}{x} - \frac{a}{x(x+a)} - \frac{\log x - a}{x^2} \\
&\approx \frac{\log(x+a) - a}{x}.
\end{align}
$$
We may then conclude from $(5)$ that
$$
W(e^{x+a}) \approx x \left(1 - \frac{\log x - a}{x+1}\right)
$$
for $x$ large and $a \ll x$.
The branches of Lambert W are the local inverses of the Elementary function $f$ with $f(z)=ze^z$, $z \in \mathbb{C}$.
The incomprehensibly unfortunately hardly noticed theorem of Joseph Fels Ritt in Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90 answers which kinds of Elementary functions can have an inverse which is an Elementary function. It is also proved in Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J. Math 101 (1979) (4) 743-759.
Ritt, J. F.: Integration in finite terms. Liouville's theory of elementary methods. Columbia University Press, New York, 1948
The non-elementarity of LambertW was already proved by Liouville in
Liouville, J.: Mémoire sur la classification des transcendantes et sur l'impossibilité d'exprimer les racines de certaines équations en fonction finie explicite des coefficients. Journal de mathématiques pures et appliquées 2 (1837) 56–105, 3 (1838) 5233–547
It is also proved in
Rosenlicht, M.: On the explicit solvability of certain transcendental equations. Publications mathématiques de l'IHÉS 36 (1969) 15-22
and in
Bronstein, M.; Corless, R. M.; Davenport, J. H., Jeffrey, D. J.: Algebraic properties of the Lambert W Function from a result of Rosenlicht and of Liouville. Integral Transforms and Special Functions 19 (2008) (10) 709-712.
Ritt's theorem shows that no antiderivatives, no differentiation and no differential fields are needed for defining the Elementary functions.
Best Answer
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,
$$x=+\sqrt{\frac{W(-\ln(2)2^{-2k-1})}{-\ln(2)}}$$