My answer is for real solutions.
$$(2x)^x=14+x^x$$
We see, this equation is a polynomial equation of more than one algebraically independent monomials ($(2x)^x,x^x$) and with no univariate factor. We therefore don't know how to rearrange the equation for $x$ by applying only finite numbers of elementary functions (elementary operations) we can read from the equation.
But we can bring the equation into a form for applying Hyper Lambert W.
$$(2x)^x-x^x=14$$
$$2^xx^x-x^x=14$$
$$(2^x-1)x^x=14$$
$$(e^{\ln(2)x}-1)e^{\ln(x)x}=14$$
$x\to\frac{\ln(1+t)}{\ln(2)}$:
$$t(1+t)^{\frac{\ln\left(\frac{\ln(1+t)}{\ln(2)}\right)}{\ln(2)}}=14$$
$$te^{\frac{\ln(1+t)\ln\left(\frac{\ln(1+t)}{\ln(2)}\right)}{\ln(2)}}=14$$
$$G\left(\left[\frac{\ln(1+t)\ln\left(\frac{\ln(1+t)}{\ln(2)}\right)}{\ln(2)}e^{-t}\right],t\right)=14$$
$$t=HW\left(\frac{\ln(1+t)\ln\left(\frac{\ln(1+t)}{\ln(2)}\right)}{\ln(2)}e^{-t},14\right)$$
$$x=\frac{\ln\left(1+HW\left(\frac{\ln(1+t)\ln\left(\frac{\ln(1+t)}{\ln(2)}\right)}{\ln(2)}e^{-t},14\right)\right)}{\ln(2)}$$
So we have a closed form for $x$, and the representations of Hyper Lambert W give some hints for calculating $x$.
Galidakis, I. N.: On solving the p-th complex auxiliary equation $f^{(p)}(z)=z$. Complex Variables 50 (2005) (13) 977-997
Galidakis, I. N.: On some applications of the generalized hyper-Lambert functions. Complex Variables and Elliptic Equations 52 (2007) (12) 1101-1119
