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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.


marked as duplicate by Dietrich Burde, Xander Henderson, jvdhooft, Arnaud Mortier, user99914 Jun 13 '18 at 20:53

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  • $\begingroup$ I'm not sure this is a duplicate of the questions listed above, except for the simplest case of $x^x=a$. OP seems to be interested in the algebraic properties of "tetration" in general. (That said, there may be duplicate questions about tetration. After all, there's a tetration tag.) $\endgroup$ – Blue Jun 12 '18 at 15:38
  • $\begingroup$ For the most basic case of $x^x=a$, you can use the Lambert-W function. You will find that $x=e^{W(\ln a)}$. $\endgroup$ – Zachary Jun 12 '18 at 15:42

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.


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