Proving that $\ln ^3|x|=x$ has exactly 3 real solutions 
Prove that
  $$\ln ^3|x|=x$$
  Has exactly 3 real solutions.

So far my idea is to look separately at $x>0$ and $x<0$ but I'm still stuck with the former.
Let $$f(x)=\ln^3|x|-x$$
Assuming $x>0$, as $\lim_{x\to0^+}=-\infty$ and $\lim_{x\to\infty}=-\infty$, and as $f(e^2)>0$ we know by intermediate value theorem that we have at least two positive solutions.
Similarly assuming $x<0$, as $\lim_{x\to0^-}=-\infty$ and $\lim_{x\to-\infty}=\infty$ There is at least one solutions.
But I am stuck at proving there are at most two solutions positive solutions and one negative, as, for example assuming $x>0$ when looking at the derivative.
$$f'(x)=\frac{3\ln^2(x)}{x}-1$$
It has a minimum at $x=1$, maximum at $x=e^2$, and by the limits at $0$ and infinity it turns out to have 3 solutions, so I can't use Mean Value theorem to prove it, and I don't really know of any other way of doing it.
What other methods can be used to prove the uniqueness of solutions here?
 A: For $x<0$, you have $y=\ln|x|$ a monotonously decreasing function that visits all reals. It obviously intersects $y=x$, which is monotonously increasing, once and only once.
For $x>0$, you can restrict to $x>1$ (because of the signs). I think this will help you restrict the number of solutions. On the interval $x\in(1,e^2)$, the function $\ln^3 x-x$ has at least one zero, because it is negative on the left and positive on the right boundary. Because it is also convex on this interval, it cannot have more than one zero. The same argument works for $x\in(e^2,\infty)$.
A: HINT:
You want to understand how $f(x) = \log^3 x - x$  (x>0) varies, look at its derivative $-1 + \frac{3 \log^2 x}{x}$. The sign is not clear, so look at the second derivative $\frac{3 (\log x -2) \log x}{x^2}$ whose sign is easy to understand. 
OK, so we see that the zeroes of $f''$ are $1$ and $e^2$. On the intervals $(0,1)$, $(1,e^2)$, $(e^2, \infty)$ the signs of $f''$ are $-$,$+$, $-$. Therefore, on the intervals $(0, 1]$, $[1,e^2]$ respectively $[e^2, \infty)$ the function $f'$ is strictly decreasing, increasing, respectively decreasing. The values of $f'$ at $1$ and $e^2$ are $-1<0$ and $-1 + \frac{12}{e^2} > 0$. Moreover, $\lim_{x \to 0} f'(x) = \infty$ and $\lim_{x\to \infty} f'(x)= -1$. Therefore, $f'$ has exactly three roots, one in the interval $(0,1)$ , one in the interval $(1,e^2)$ and one in the interval $(e^2, \infty)$. Let's denote them by $c_1$, $c_2$, and $c_3$. So $f$ is strictly increasing on $(0, c_1]$, decreasing on $[c_1, c_2]$, increasing on $[c_2, c_3]$ and decreasing on $[c_3, \infty)$.  One checks that $f(c_1) < 0$, $f(c_2) < 0$ and $f(c_3) >0$. Indeed, on $(0,1)$ $f$ is clearly negative, so $f(c_1) < 0$. Then $f(c_2) < f(c_1) < 0$. On $[c_2, c_3]$ $f$ is strictly increasing so $f(c_3) > f(e^2) > 0$. Moreover, $\lim_{x \to \infty} f(x)= - \infty$. Now it's clear that $f$ has exactly two roots, one in $(c_2, c_3)$ and one in $(c_3, \infty)$. 
Plot $f$, $f'$ and $f''$, once closer to $0$, and once around $100$ and you see all of the above.  
