# Solve the tower equation: $\displaystyle x^{x^{x^{x+1}+x+1}}=2$

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?

• Numerical Solution : $x=1.28486857142857$ gives output $1.99998789347940$ while $x=1.28487142857142$ gives output $2.00001726575946$
– Prem
Commented Jun 10, 2023 at 19:18
• How is \displaystyle allowed in the title? Commented Jun 11, 2023 at 13:37

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

$$\boxed{x=e^{W(W(W(\ln2)))}}$$

• Since "can be solved mathematically" doesn't have a clear definition, this answer should be the best to expect. Commented Jun 10, 2023 at 19:21