Do you know Soumava's Algorithm?

This can be solved using the Newton-Raphson method but first we need to set up our function and establish the number of roots. First, note that: $\begin{aligned} x^y &= 43 \\ y\ln(x) &= \ln(43) \\ y &= \frac{\ln(43)}{\ln(x)}. \end{aligned}$ Substituting this back into the first equation gives: $y^x = \left(\frac{\ln(43)}{\ln(x)}\right)^x.$ The function that we want to find the root of is: $\begin{aligned} f(x) &= \left(\frac{\ln(43)}{\ln(x)}\right)^x - 43 \\ &= e^{x\ln\left(\frac{\ln(43)}{\ln(x)}\right)} - 43 \end{aligned}$

We now need to establish the number of roots that this function has. It is actually simpler to consider $x^y = y^x$ . We can rearrange this as: $\frac{\ln(y)}{y} = \frac{\ln(x)}{x}$ Clearly $x = y$ is always a solution to this but since we need $x$ and $y$ to be distinct we are only interested in other solutions. Fortunately $g(x) = \ln(x)/x$ is not injective. We can sketch it by noting the following facts: $\begin{aligned} \lim_{x \to 0} g(x) &= -\infty \\ g(1) &= 0 \\ \lim_{x \to \infty} g(x) &= 0 \\ g^\prime(x) &= \frac{1 - \ln(x)}{x^2} \\ g^\prime(e) &= 0 \end{aligned}$ Also, note that $x = e$ is the only turning point. Thus $g(x)$ looks like this: So $g(x) = k$ has precisely two solutions for any $k \in (0, 1/e)$ , one solution in the range $(1, e)$ and the other in the range $(e, \infty)$ .

We cannot directly map this back onto the original problem since we also need $\ln(x)/x = \ln(43)/xy$ . What we can infer, though, is that the solution set to $x^y = y^x$ consists of the line $y = x$ , as well as a smooth curve from the region $(1,e) \times (e, \infty)$ , through the point $(e,e)$ , to the region $(e,\infty) \times (1,e)$ .

The solutions to the original problem correspond to the intersection between this curve and $y = \ln(43)/\ln(x)$ . This will have either one or three solutions, depending on whether it intersects off the line $y = x$ . Therefore, we will search for a solution to $f(x) = 0$ near $x = 1.1$ .

Starting at $x_0 = 1.1$ , and applying the Newton-Raphson method, after 5 iterations we have $x_5 = 1.156752515$ . The corresponding value for $y$ is $y_5 = \ln(43)/\ln(x_5) = 25.82948719$ . The required solution is then $\lfloor 100(x + y) \rfloor = \boxed{2698}$ .