Solve $x^x=2x$ where $x\in\mathbb C$.

$$x^x=2x$$

$$=e^{x\ln x}=e^{(a+bi)(\ln|a+bi|+i(\arg(a+bi)+2k\pi))}$$

$$=e^{a\ln|a+bi|-b(\arg(a+bi)+2k\pi)}\cdot e^{bi\ln|a+bi|+ai(\arg(a+bi)+2k\pi)}$$

$$\implies {e^{a\ln|a+bi|}\over e^{b(\arg(a+bi)+2k\pi)}}=2|a+bi|$$

$$\text{and}\qquad e^{bi\ln|a+bi|+ai(\arg(a+bi)+2k\pi)}={a+bi\over|a+bi|}$$

$$\implies (a^2+b^2)^{a-1\over 2}=2e^{b\arg(a+bi)+2bk\pi}\tag 1$$

$$\text{and}\qquad\text{cis}(b\ln\sqrt{a^2+b^2}+a\arg(a+bi)+2ak\pi)=\text{cis}(\arg(a+bi)+2l\pi)\tag 2$$

$$(2)\implies \frac b2\ln(a^2+b^2)+(a-1)\arg(a+bi)=2l\pi-2ak\pi\tag 3$$

This is almost to the point of evaluating all-real numbers. Fortunately, many graphing programs will evaluate $\arctan(\tfrac ba)$ using all branches and correctly evaluating at e.g. $x=0$ such that $\arg(a+bi)$ can be graphed as $\arctan(\tfrac ba)$. Here are some graphs with $a$ along the $x$-axis and $b$ along the $y$-axis:

As demonstrated in the pictures, there are the three real solutions at $x\approx -0.658, 0.346, 2$, and there are many complex solutions, one of which is near the location $3+3i$.

Also, we have two equations to work with, so we can continue the direct analysis:

$$(1)\implies a^2+b^2=(2e^{b\arg(a+bi)+2bk\pi})^{2\over a-1}$$

Plugging into $(2)$:

$$\frac b2\ln((2e^{b\arg(a+bi)+2bk\pi})^{2\over a-1})+(a-1)\arg(a+bi)=2l\pi-2ak\pi$$

$$=\frac b{a-1}(b\arg(a+bi)+2bk\pi)\ln2+(a-1)\arg(a+bi)$$

$$=\frac {b^2\ln 2+(a-1)^2}{a-1}\arg(a+bi)+\frac{2b^2k\pi\ln 2}{a-1}$$

Unfortunately, this doesn't lend itself to any further breakdown that I am aware of. The graph is not easy to create, since the branches appear to be a near-full covering of 2D space, and even setting e.g. $k=1$ only makes for confusing graphs that do not correspond to the solutions identified above.