We are struggling to solve a crazy looking equation below:
$$\displaystyle x^{x^{x^{x+1}+x+1}}=2$$
Approximate numerical values are unfortunately not a solution. Wolfram Alpha offers only the following approximate solution:
$$x\approx 1.28486974908322\dots$$
A quote from the author: " The equation can be solved mathematically " .
This equation is not a homework and it seems hard to solve. At least it seems that way to me. However, I know the solution of an equation similar to this equation .
$$x^{x^{x^{x^{x^{\dots}}}}}=2$$
So, we have
$$x^2=2$$
which implies $x=\sqrt {2}$ .
My opinion
Probably $x$ is an algebraic number. For example, $x=\sqrt {\frac {3}{2}}$. However, it is difficult to find.
My questions
Do you think we can establish a similarity between the solutions of these equations?
Wolfram Alpha cannot solve the equation for the exact value of $x$ . Can other computer software solve the equation?
Best Answer
First, we need to denote the solution to the following basic equation $$t^t=2\Longrightarrow t=e^{W(\ln2)}\tag{1}$$
where $W$ is Lambert-W function, and you can find here. Now, we are ready to solve it. $$L=x^{x^{x^{x+1}+x+1}}=\left(x\right)^{x\cdot x^x\cdot x^{x^{x+1}}}=\left(x^x\right)^{x^x\cdot x^{x^{x+1}}}$$
Let $x^x=y$
$$L=\left(y\right)^{y\cdot x^{y\cdot x}}=y^{y\cdot (x^{x})^y}=y^{y\cdot y^y}$$
Let $t=y^y$
$$L=t^t=2$$
Now, use (1), we get
$$\boxed{x=e^{W(W(W(\ln2)))}}$$
Numericall verify:
and