Solving an equation that contains a logarithm I have the follwing equation:
$$y=\frac 1 4x^2 -\frac 1 2 \ln{x}$$
How can $x$ be expressed in terms of $y$?
 A: We have 
$$y=\frac14 x^2-\frac12 \log x\tag 1$$
Multiplying $(1)$ by $-4$ yields
$$\begin{align}
-4y&=- x^2+2 \log x\\\\
&=-x^2+\log x^2 \tag 2\\\\
\end{align}$$
Now, exponentiating both sides of $(2)$ and multiplying by $-1$ reveals that
$$-e^{-4y}=-x^2e^{-x^2} \tag 3$$
from which we find that 
$$x=\sqrt{-W(-e^{-4y})}$$
where $W(z)$ is the Lambert W Function.  Note we rejected negative values of $x$ since in $(1)$ we have $\log x$.
We must have $W\le 0$ and $-e^{-4y}\ge -e^{-1}\implies y\ge \frac14$.
A: As said in comments, the solution of equation $$\frac 1 4x^2 -\frac 1 2 \ln{x}-y=0$$ cannot be expressed in terms of elementary functions.
However, there are explicit solutions in terms of Lambert function. Chenging variable $x^2=t$, the equation write $$t-\log(t)-4y=0$$ and the solution is $$t=-W\left(-e^{-4 y}\right)$$
You must notice that $y'=\frac{x}{2}-\frac{1}{2 x}$ cancels at $x=1$ and, at this point $y=\frac 14$ and the second derivative is always positive. So, if $y<\frac 14$, there is no real root; if $y=\frac 14$, the only root is $x=1$ and if $y>\frac 14$, there are two real roots (one between $0$ and $1$, another larger than $1$).
