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How to solve this equation for $x$ where $a>0$? The exponent tower goes on forever:

$$a=x^{x^{x^{.^{.^{.}}}}}$$

My Calculus book gives the following reasoning:

$$ln(a)=x^{x^{x^{.^{.^{.}}}}}ln(x)=a\,ln(x)$$

To conclude that: $$x=a^\frac{1}{a}$$

Why is this correct?

GambitSquared
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    The meaning of the expression on $\text{RHS}$ of the equation eludes me. That is, how is the expression $x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}$ defined? Clearly it cannot be in the obvious way -- i.e., left to right, for that clearly diverges for any $x>1.$ So, what does this expression define? – Allawonder Apr 30 '19 at 03:14
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    Wikipedia says something useful about the convergence. We need to have $x$ in the range $[e^{-e},e^{1/e}]$ for the power tower to converge. – Jyrki Lahtonen Apr 30 '19 at 03:36
  • According to the comments posted by the author, it seem that he/she is confused with the existence of the limitation, namely, $a$, but the OP says nothing about the existence of $a$. – Zongxiang Yi Apr 30 '19 at 08:05

5 Answers5

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Equation: $$a = {x^{x^{x^{.^{.^{.}}}}}}$$

Take $ln$ of both sides, then use the power rule in right and plug $"a"$ in ${x^{x^{x^{.^{.^{.}}}}}}$:

$$ln(a) = ln({x^{x^{x^{.^{.^{.}}}}}})$$ $$ln(a) = {x^{x^{x^{.^{.^{.}}}}}} \cdot ln(x)$$ $$ln(a) = a \cdot ln(x)$$

Divide both sides by "$a"$, then rearrange left side for using power rule afterwards: $$\frac{ln(a)}{a} = ln(x)$$ $$\frac{1}{a}ln(a) = ln(x)$$

Use power rule in left, then take $exp$ of both sides and lastly, rewrite left side as a root: $$ln(a^\frac{1}{a}) = ln(x)$$ $$a^\frac{1}{a} = x$$ $$\sqrt [a]{a} = x$$

Kayrab Cebll
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Re-write the first equation in your question as

$$a=x^a$$

and the result follows.

Matt L.
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    I don't think it's that simple. When we solve an equation we want to find a necessary and sufficient condition on the independent variable(s), but this only gives a necessary condition. In other words: you have shown $$x^{x^{x^{.^{.^{.}}}}}=a\implies x^a=a\implies x=a^{1/a}, $$ but we also need to show $$x=a^{1/a}\implies x^{x^{x^{.^{.^{.}}}}}=a. $$ – Oskar Henriksson Nov 06 '17 at 05:07
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The key point is to notice that $\lim_{n\to \infty}{a_n}=\lim_{n\to \infty}{a_{n+k}}$ for any $k\in \mathbb{Z}$. It means that the limitation does not depend on the first $k$ items. Define a sequence: $a_n=x^{a_{n-1}}$ for $n\ge 2$ and $a_n=x$ for $n=1$.

So we can say $a=x^a$, where the $a$ in the left hand side is $\lim_{n\to \infty}{a_n}$ and the $a$ in the right hand side is $\lim_{n\to \infty}{a_{n+1}}$.

Finally, we have $x=a^{\frac{1}{a}}$.

Zongxiang Yi
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    The answer suffers from the weakness exposed in Oskar Henriksson's comment under Matt L's answer. You have not justified that the limit exists. Granted, the OP's textbook has the same weakness (and makes the said textbook dubious). – Jyrki Lahtonen Apr 30 '19 at 03:25
  • @JyrkiLahtonen Obviously, if $x=1$ then the limitation exists and $a=1$. It means that there exists $D\subset \mathbb{R}$, where the limitation exists. So is your question trying to show that $D=\mathbb{R}$? Since the OP says $a>0$, it tells me that the limitation must exist. Otherwise , how can we say the limitation is geater than 0. – Zongxiang Yi Apr 30 '19 at 03:36
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    How does $a>0$ tell you that the limit must exist? – Jyrki Lahtonen Apr 30 '19 at 03:43
  • $>$ is a binary relation defined on $\mathbb{R}$ So $a \in \mathbb{R}$ and the limitation exists. – Zongxiang Yi Apr 30 '19 at 03:45
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    Does not follow. YOU have to prove that $\lim_{n\to\infty}a_n$ exists before you can conclude that $a=x^a$. Otherwise you can do things like the following. Let $$S=1-1+1-1+1-1+1-1+\cdots.$$ Then $$1-S=1-(1-1+1-1+1-\cdots)=1-1+1-1+1-1+\cdots=S,$$ and then conclude that we must have $1-S=S$ so $S=1/2$. The error was that by calling that sum $S$ we silently assumed it exists. – Jyrki Lahtonen Apr 30 '19 at 03:57
  • Note that the existence of this limitation depend on $x$, For example, if $x=1$, then $a$ exists and if $x=2$, then $a$ does not exist. So you book assumes that the limitation exists and try to compute the limitation. It seems that you want to know when the limitation exists, w.r.t. $x$. – Zongxiang Yi Apr 30 '19 at 04:04
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$x^{x^{x^{x^{x^\ldots}}}} = a$

This can easily be rewritten as $x^a = a$, which gives the root $x = \sqrt[a]{a}$

There is, however, one problem: $2=\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^\ldots}}}} = \sqrt[4]{4}^{\sqrt[4]{4}^{\sqrt[4]{4}^{\sqrt[4]{4}^{\sqrt[4]{4}^\ldots}}}} = 4$

Jack Smith
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Your book is right. Don't consider that $x^n$ is infinite. It is finite here in this case, and equal to $a$ as given in the question. Your book has a very good explanatory logic.

SchrodingersCat
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    I don' agree. The only thing the book seems to show (as presented in the OP) is that if there is a solution, it has to be $x=a^{1/a}$. But how do you know that there actually is any solution at all? This is an interesting example of what can happen if one fortgets to check that: https://math.stackexchange.com/questions/492109/what-is-wrong-with-this-funny-proof-that-2-4-using-infinite-exponentiation. – Oskar Henriksson Nov 06 '17 at 10:48