How can I solve the equation for $x$ when $x^x=2$ or any other constant? And is solving $x^{x^x}=a$ or $x^{x^{x^x}}=a$ or equations such as these even possible? What are these equations even called?
And what about the following? $$x^{(x-1)^{(x-2)^{\dots^{3^{2^{1}}}}}}=k$$ or $$a\;x@n+b\;x@(n-1)+c\;x@(n-2)+\cdots=0$$ where "$@$" is an operator I made to indicate the number of exponents, as $x@3:=x^{x^x}$.
I have already tried taking logarithms, using taylor series, etc.
Let us first make the substitution $x = e^t$. $$a = \left(e^t\right)^{e^t}$$ $$a = e^{te^t}$$ $$\ln a = te^t$$ This is now of the form $y = xe^x$. The inverse of this equation is known as the Lambert W function. This means that $$t = W(\ln a)$$ $$x = e^{W(\ln a)}$$ After this point, you must evaluate on a case by case basis. You could either look up specific values of Lambert W or you could find the values using either Newton's Method or Halley's Method.