Complex Chicanery

I came across an interesting video on YouTube, uploaded today on a channel named Higher Mathematics that posed the following problem: $$\text{Solve for } x:\\1^x=5$$ Clearly the problem has no solution amongst the real numbers but there turns out to be an infinity of solutions once we introduce complex numbers. The solution (with $k=1,2,3 \dots$ ) then unfolds: $$\begin{align} e^{i \theta} &= \cos{\theta}+i \sin{\theta}\\e^{i 2k \pi} &= \cos{2k \pi}+i \sin{2k \pi}  \\&=1 \\ \therefore (e^{i 2k \pi)})^x &= 5\\ e^{i 2k \pi x} &= 5 \\\ln {e^{i 2k \pi x} }&= \ln{5} \\i 2k \pi x &= \ln{5} \\x &= \frac{ \ln{5}}{i 2k \pi} \\\ &=\frac{-i \ln{5}}{2k \pi}\end{align}$$ The complex numbers for values of x given by $k=1, 2,3$ are shown in Figure 1. They all map to the point $5+0i$.

Figure 1

Thus we have a function $f(x)$ with its domain D being the countably infinite set of points: $$\left( \Big( 0,\dfrac{-i \ln{5}} {2k \pi} \Big) \text{ where k=1, 2, 3, ...} \right)$$  that maps all these points to $(5, 0i)$ so that: $$f(x) \rightarrow (5, 0i) \text{ for all x in D}$$

*********************

Let's look now at a slightly different problem: $$2^x=x$$ Right from the start, it can be said once again that there are no real solutions to this equality. This is clear once we graph the line $y=2^x$ and $y=x$ where we see that there are no points of intersection. See Figure 2.

Figure 2

This problem was posed on the same YouTube channel mentioned earlier. Here is a link to the video. So how do we find if any complex numbers satisfy this equality? Firstly, we take the natural logarithms of both sides and proceed from there: $$\begin{align} 2^x &= x \\ \ln{2^x} &= \ln{x} \\ x \ln{2} &= \ln{x} \\ \frac{\ln{x}}{x} &= \ln{2} \\ \frac{\ln{x}}{e^{\, \ln{x}} }&= \ln{2} \\ \ln{x} \, {e^{\, -\ln{x}} } &= \ln{2} \\ -\ln{x} \, {e^{\, -\ln{x}} } &= -\ln{2} \\ W(-\ln{x} \, {e^{\, -\ln{x}} } ) &= W(-\ln{2}) \\ -\ln{x} &= W(-\ln{2} ) \\ \ln{x} &= -W(-\ln{2}) \\ x &= e^{-W(-\ln{2})} \\ x &= \frac{1}{e^{W(-\ln{2})}} \\ x &\approx -0.37927 - 0.72087 i\end{align}$$ where W is the Lambert W function that I've written about in previous posts. See The Omega Constant and the Lambert W Function (June 24th 2020) and More on the Lambert W Function (February 16th 2021). To evaluate $x$ using Wolfram Alpha the command ProductLog[-ln(2)] needs to be used.