Don't be fooled!

Here is the best way to SEE why this happens (rather than just verifying it analytically):

(a) Draw the graphs of the function $(\sqrt{2})^x=2^{x/2}$ and $x$ on the same axes, intersecting at $(2,2)$ and $(4,4)$ .

(b) Now draw the cobweb starting at $x=\sqrt{2}$ and see how it zig-zags towards the point $(2,2)$ . The cobweb method is explained step-by-step here .

I think the function $f(x) = a^x-x$ has multiple roots and the iteration used in the exercise converges against them from below and you therefore end up with the wrong solution in this case(See this https://brilliant.org/problems/more-fun-with-power-towers/?group=yvVpX9gRy9Vh&ref_id=1116352)

Not in the function $f(x)=x^{x^{x^{x^{...}}}}$ , but in the relation $y=x^{x^{x^{x^{...}}}}$ , and we can realize that is the same that $y=x^y \rightarrow \sqrt[y]{y}=x$ . With this we can realize how its graph is (If you know it). The maximum value possible for $x$ is $\sqrt[e]{e}$ and its relative to the point $(\sqrt[e]{e},e)$ which belongs to the graph, after that point the value of $x$ begins to return slowly, converging to $x=1$ , while $y$ is still going up forever, so I think $4$ is right in the range the positive solution can actually be $\sqrt{2}$ , but all we know that the infinite power of $\sqrt{2}$ is $2$ but not $4$ .... I think this is maybe a case similar to the case of $sqrt{1}$ and $x^2=1$ . We should know that $\sqrt{1}=1$ and the solutions for $x^2=1$ are $x=1$ or $x=-1$ , but this doesn't mean that $\sqrt{1}$ may be $1$ or $-1$ . In this problem we are asking if is there any positive solution for $x^{x^{x^{x^{...}}}}=4$ and this is the same that asking if the relation $y=x^{x^{x^{x^{...}}}}$ has a positive $a$ such that $(a,4)$ belongs to its graph, and the point $\sqrt{2},4$ does......

Well, that's the question I'm asking: Is 4 in the range? My answer is "no".