# What is $x$ if $x^{x^x} = {(1/2)}^{\sqrt 2}$?

What is $$x$$ if $$x^{x^x} = {(1/2)}^{\sqrt 2}$$?

Answer provided more or less like this : \begin{align}{\left(\frac 12\right)}^{\sqrt 2}&=\frac1{2^{\sqrt 2}}\\ 2^{\sqrt 2} &= 2^{(2^{(2/4)})} \\ &= 2^{(4^{(1/4)})}\\ & = 4^{1/2(4^{(1/4)})} \\ &= 4^{2^{-1}(4^{(1/4)})} \\ &= 4^{4^{-1/2}(4^{(1/4)})}\\ &= 4^{(4^{(-1/4)})}\\ &= 4^{({1/4}^{(1/4)})} \\ {\left(\frac 12\right)}^{\sqrt 2} &= \frac1{2^{\sqrt 2}} \\ &= \frac1{4^{({1/4}^{(1/4)})}} \\ &= {\left(\frac14\right)}^{({1/4}^{(1/4)})} \\ x&=\frac 14\end{align}

Is there more elegant way that show $$x=\frac 14$$ the only answer?

Edit : source https://youtu.be/d-E5isaIDTA

• well I think your way is a good one, since for another ways you have to you numerical algorithms which are not as elegant as yours – user715522 Feb 5 at 16:17
• good solution, indeed. $\to +1$ – Claude Leibovici Feb 5 at 16:35
• It's not the only answer, if you consider the complex plane. Using numerical methods you find x=2.3528266687 - i * 0.746689357999 works. – skbmoore Feb 5 at 16:35
• @skbmoore. There are so many solutions in the complex domain ! – Claude Leibovici Feb 5 at 16:40
• The OP is almost certainly interested only in positive real values for $x$. But their approach only shows that $x=1/4$ is a solution, not that it's the only solution. I for one am hard pressed to think of a pre-calculus way of ruling out the existence of other (real) solutions. (Indeed, even a calculus-based proof looks like it'll be a little complicated.) – Barry Cipra Feb 5 at 16:47

Here is a calculus-based proof that $$x=1/4$$ is the only positive real solution to the equation $$x^{x^x}=(1/2)^\sqrt2$$.

It suffices to show that $$x^{x^x}$$ is increasing for all $$x\gt0$$. This is clear for $$x\ge1$$, so it remains to consider what happens for $$0\lt x\lt1$$. It's convenient to let $$x=e^{-u}$$ with $$u\gt0$$, and, after taking logarithms (twice), show that $$f(u)=\ln u-ue^{-u}$$ is increasing (from $$-\infty$$ as $$u\to0$$ to $$\infty$$ as $$u\to\infty$$). For this we need to show that

$$f'(u)={1\over u}-(1-u)e^{-u}={1-u(1-u)e^{-u}\over u}\ge0$$

for all $$u\gt0$$. But this is clear since $$u\gt0$$ implies $$e^{-u}\lt1$$, hence

$$u(1-u)e^{-u}\lt u(1-u)\le{1\over4}\lt1$$

(the maximum for $$u(1-u)$$ occurring at $$u=1/2$$).

If there is a non-calculus proof that $$x^{x^x}$$ is increasing, I'd be keen to see it.

Here is a simpler proof of the fact that $$f$$ is strictly increasing (and therefore that the equation has at most one solution in the positive reals):

The function $$f(x) = x^{x^x}$$ can be expressed as $$f(x) = e(x, e(x, x))$$ where $$e(x, y) = x^y$$. If the function $$e$$ is strictly increasing in each argument, then so if $$f$$:

$$x < y$$ implies $$f(x) = e(x, e(x, x)) < e(x, e(x, y)) < e(x, e(y, y)) < e(y, e(y, y)) = f(y)$$.

But $$e$$ being strictly increasing in each argument is equivalent to the functions $$g_c(x) = c^x$$ and $$h_c(x) = x^c$$ being increasing for each positive real $$c$$. This can reasonably be taken as given.

(If you wanted to prove that $$g_c$$ and $$h_c$$ are strictly increasing, it would have to be proved from the limit definition of exponentiation. A calculus proof of this fact would miss the point that the fact that these functions are strictly increasing is more basic than whatever rule you might have for computing their derivatives.)

• I don't think the inequality $e(x,e(x,y))\lt e(x,e(y,y))$ is true when $0\lt x\lt 1$. E.g., $e(1/2,e(1/2,1))=(1/2)^{1/2}=\sqrt{1/2}\gt e(1/2,e(1,1))=(1/2)^1=1/2$. – Barry Cipra Mar 18 at 18:57
• You are absolutely right, I was being sloppy. The function e is decreasing in the second argument on the relevant interval, not increasing. – Pilcrow Mar 19 at 20:48