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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?

wythagoras
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Nescio
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  • Not tried, but if you are stuck because you can't really know enough from just the behavior of $f^\prime$, try to differentiate twice to get a better picture of the variations of the function. – Clement C. Jun 29 '15 at 07:20
  • @ClementC. I actually tried that, from the second derivative I found out the minimum and maximum of $f'$ Now I could go to a third derivative as it is already pretty simple, but it's so far I have no idea how to use it in my original problem – Nescio Jun 29 '15 at 07:22
  • Interesting question, the behavior of $\ln(x)^r-x$ for $r\in(2,4)$ is actually somewhat complicated. I'd like to know at which $r$ the number of $x$-intersections in which interval changes. PS: Accepting an answer fast, e.g. within a time where half of the world is sleeping, doesn't help you or people who potentially want to contribute. – Nikolaj-K Jun 29 '15 at 08:12
  • @NikolajK Correct me if I'm wrong, but I am supposed to accept an answer once I understood the solution to my question, am I not? I really wouldn't mind leaving questions open for longer but it does seem like I'm supposed to accept it when I solved it to prevent people putting in effort for nothing instead of answering still open questions. – Nescio Jun 29 '15 at 08:43
  • @Nescio: I'm not sure. You left an hour to answer before it was done - maybe that's the best way if you're sure you're done with the problem. But I think otherwise you may get a more detailed response. I leave question open for one day on principle, just so that the standard userbase gets to see it. If the question pops up with a the green being solved mark, I assume people will tend to not put more though into it. I don't know how it affects the stream of people visiting would criticise the answer or point out errors. – Nikolaj-K Jun 29 '15 at 08:50

2 Answers2

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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 $x\in(e^2,\infty)$.

orion
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  • Actually just the fact that the function must be negative on $(0,1)$ is the obvious thing I was missing ><. Thank you – Nescio Jun 29 '15 at 08:06
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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.

orangeskid
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