Repeating Power

Relevant wiki: Nested Functions

As Arthur and Andrew point out, if there is a solution at all, it must be $x=\ln(10)-10$ . However, we must examine whether the power tower with $x=\ln(10)-10$ does indeed converge to 10. It will turn out that it doesn't, so that this problem has no solutions .

It is convenient to introduce the iteration function $f(t)=e^{x+t}=e^{\ln(10)-10+t}$ , an exponential growth function. Now we can write our iteration simply as $a_{n+1}=f(a_n)$ , with the seed $a_0=0$ , and we are asked to find out whether $\lim_{n\to\infty}a_n=10$ .

It turns out that $f(t)$ has two fixed points , where $f(t)=t$ , namely, $b=-W\left(-\frac{10}{e^{10}}\right)\approx 0.0004542$ and $c=10$ .

I suggest that the reader graphs the functions $f(t)$ and $t$ on the same axes to get a sense for what's going on here. Draw a cobweb for various initial values if you are familiar with the concept...that helps a lot in really understanding the issue of convergence.

Now we can show by induction that $a_n<b$ for all $n$ . We know that $a_0=0<b$ . Assuming $a_{n}<b$ we conclude that $a_{n+1}=f(a_n)<f(b)=b$ since $f(t)$ is increasing.

Since $a_n<b <0.0005$ for all $n$ , the limit of $a_n$ cannot be 10. It turns out that the limit is actually $b$ , but that observation is not crucial for our solution (left as an exercise).

Let me mention that this convergence behavior is closely related to the fact that $f'(b)=f(b)=b<1$ while $f'(10)=f(10)=10>1$ : The equilibrium $b$ is stable while $c=10$ isn't.

I approached this slightly differently than the other solution, so I may as well post it:

${ e }^{ { x+e }^{ { x+e }^{ { x+e }^{ x+e } } } }=10\\ { e }^{ x+10 }=10\\ x+10=\ln { (10) } \\ x=\ln { (10) } -10\\ x\cong -7.697$

$e^{x + 10} = 10 \\ x \approx -7.697$

but, as $x \approx -7.697$ , the value of the nested function above is approaching 0 which is not equal to 10.

Proof:

by just trying to substitute and observe the obtained results as $x \approx -7.697$ ,

as x < 0 < e , x + e will turn out negative.

$0 < e^{x+e} = e^{- something} < 1 \\ 0 < e^{x+e^{x+e}} = e^{x+ e^{x + e}} = e^{x + e^{ -something}} < 1 \\ \vdots$

it completely suffices that as $x \approx -7.697$ , the infinite nested function $e^{x+e^{x+e^{x+e^{x+\cdot^{\cdot^\cdot}}}}}$ will approach zero and not equal 10.

$\therefore$ there is no such x exist.

$e^{x+e^{x+e^{x+e^{x+e^{x+e^{x+e^{x+...}}}}}}}=10$

Start by applying the natural log on both sides:

$x+e^{x+e^{x+e^{x+e^{x+e^{x+e^{x+e^{x+...}}}}}}}=\ln(10)$

The part that comes after the x+ is the same as what we started with, so we can replace that with 10.

$x+10=\ln(10)$

Solve for x.

$x=\ln(10)-10=-7.697$