$\frac{x^x + x^x + x^x }{x + x + x} = 9$, $x^{-x} = ?$

Question

$$\frac{x^x + x^x + x^x }{x + x + x} = 9$$ $$x^{-x} = ?$$

This problem seems very complex to me. I've found the answer as $\frac {1}{27}$. However, I don't think that I've found the correct answer. Can you assist?

Answer 1

Hint:

$$9 = \frac{x^x+x^x+x^x}{x+x+x} = \frac{3x^x}{3x} = \frac{x^x}{x} = 9.$$

I don't know of a way to solve this analytically, but a guess of $x=3$ shouldn't be too hard to make from here.

Note: There is actually another solution at $x\approx 0.0895224$ which gives $x^{-x}\approx 1.24115$. Because of this, it is not likely than an analytical solution is possible, nor is there a unique answer to the question. It is likely that the question writer(s) intended this to be a "guess-and-check" sort of problem.

Answer 2

We have $$x^x=9x$$

Since, $f(x)=x^x$ is a convex function, we see that this equation has two roots maximum.

But, $3$ is a root and there is a root $x_1$, where $0<x_1<1$.

For $x=3$ we obtain $x^{-x}=\frac{1}{27}.$

Also, $$x_1^{-x_1}=1.24...$$

Answer 3

Note that

$$\frac{x^x + x^x + x^x }{x + x + x} =\frac{3x^x}{3x}=x^{x-1} = 9\implies x=3$$

to justify the solution and check for others we can consider

$$x^{x-1} = 9\iff \log x^{x-1} = 2\log 3\iff (x-1)\log x=2\log 3$$

and study the zeros for the function

$$f(x)=(x-1)\log x-2\log 3$$

Notably note that

$$f'(x) = -\frac1x +\log x +1=0$$

and to study this we can consider

$$g(x)=f'(x) \implies g'(x)=\frac{x+1}{x^2}>0 \quad \forall x>0$$

then f'(x) may have only one zero that is $x=1$ by inspection.

It easy to show, with a few of work by derivatives, EVT and limits that this is a negative minimum for $f(x)$ and that another (and only one) solution have to exist in the interval $x\in(0,1)$.