# Solutions of the equation $x^x=\frac{1}{256}$

Find the solutions of the equation $$x^x=\frac{1}{256}$$ I know that the function $$f(x)=x^x$$ is defined for $$x>0$$, so the solutions, if they exists, must be $$>0$$; as far as I know, this is because to maintain the formal properties of power, we must impose that $$x>0$$ or it is easy to get contradictions like $$-1=(-1)^{\frac{2}{2}}=[(-1)^2]^{\frac{1}{2}}=1^{\frac{1}{2}}=1$$.

So, since the function $$x^x$$ has an absolute minimum at $$x=\frac{1}{e}$$ and it is $$\left(\frac{1}{e}\right)^{\frac{1}{e}}>\frac{1}{256}$$, it follows that there isn't an $$x>0$$ such that $$x^x=\frac{1}{256}$$ and it follows that the equation hasn't real solutions.

However, Wolfram|Alpha says that the equation has the integer solution $$x=-4$$ and it is indeed a solution, as one can check by substitution. However, for what I said before, the function $$x^x$$ doesn't exist for $$x<0$$ and so it can't be evaluated for that value.

Why is this happening? Has it something to do with complex numbers? Or finding a solution of that equation is different to consider the function $$x^x$$ involved? Thank you.

• $x^{x}$ is defined for $x < 0$, so I don't understand the issue. $-4$ is in fact a solution as you mentioned. Your "minimum" is only a minimum for when $f$ is strictly defined for only x $> 0.$ Jan 5 at 8:27
• $x^x$ is in general not real-valued for $x<0$. However, it is real-valued whenever $x$ happens to be an even integer: If $x=-2n$, then $x^x=(-2n)^{-2n}=(2n)^{-2n}$. Jan 5 at 8:32
• math.stackexchange.com/questions/1551470/domain-of-xx Does this answer your question??
– user822140
Jan 5 at 8:32
• Mathematica "solves" with $$x=-\frac{\log 256}{W(-\log 256)}$$ which is $x\approx -1.05228+2.28451 i$ Jan 5 at 12:50

## 3 Answers

The problem is here:

the function $$x^x$$ doesn't exist for $$x < 0$$

This is wrong. Or, if you take it this way, then $$x = -4$$ cannot be and is not a solution, for then $$(-4)^{-4}$$ would have to be undefined as well.

You see, whenever you talk about functions, you cannot forget the domain - and in this case, the domain of $$x \mapsto x^x$$ for $$x$$ real is a bit tricky, and your question solution lies right in the tricky part. Everyone can agree that $$(0, \infty)$$ is part of it, but not everyone necessarily agrees on what the case is for other values of $$x$$. When you say

one can check by substitution [that it is indeed a solution]

you are presuming a larger domain than the one you just gave in order to do that "substitution", and that changes the answer to the question, and indeed it changes results like "that there is no solutions below $$x = \frac{1}{e}$$".

Now Wolfram does, indeed, presume a larger domain. In fact, it takes the domain to be all of $$\mathbb{R}$$, and moreover takes the codomain to be $$\mathbb{C}$$, so you are right that complex numbers play a role here.

Even more in fact, given that the domain is part of the definition of the function, you could say you and Wolfram are working with two different functions, that both happen to be denoted by the ambiguous notation $$x^x$$. Hence there is no surprise that there will be two different answers as to what the solution set is.

Wolfram seems to take as $$x^x = \frac{1}{4^4} = \frac{1}{(-4)^4} = (-4)^{-4}$$

Yes, this does have to do with complex numbers. As you correctly observed; blindly taking powers like $$(-1)^{\frac12}$$ will lead to seeming contradictions.

Thus in complex-analysis, we make sure to be very specific in what we mean with $$(-1)^{\frac12}$$. This is done by defining the complex log, where we need to pick a branch. This way we can consider $$x^{x}$$ on a suitable space, determined by the choice of the branch. We can then vary the branch, to check all of $$\mathbb{C}$$.

Hope this was enough to show you that the contradictions can be avoided and the $$x<0$$ cases can be considered.