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Find the sum of all real solutions for $x$ to the equation $(x^2 + 2x + 3)^{(x^2+2x+3)^{(x^2+2x+3)}} = 2012.$

I just know $x^{x^x}$ is increasing in $x$ and hence the equation has a unique solution, nut then I dont know how to move on, I also know viete' formula but I dont know if it helps here, thanks in advance.

amy tsang
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  • Where do the brackets go? Exponentiation is not associative. For example $(2^3)^4 \neq 2^{(3^4)}$. – Fly by Night Aug 22 '13 at 19:43
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    @FlybyNight: $a^{b^c}$ means $a^{(b^c)}$ by standard convention, which matches how it's entered into LaTeX. – Jonas Meyer Aug 22 '13 at 19:44
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    @JonasMeyer I can't be that standard or else I wouldn't have asked. $a^b^c$ – Fly by Night Aug 22 '13 at 19:51
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    It is standard. One justification for it is that even though exponentiation is not associative there is rarely any reason to write $(a^b)^c$ because we can just write $a^{b \cdot c}$ instead. So $a^{b^c}$ gets to have the more useful meaning. – Dan Brumleve Aug 22 '13 at 19:57
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    @FlybyNight Well, LaTeX allows for $a^{b^c}$ and ${a^b}^c$, so to talk about the standard indeed requires a good eye :) – Hagen von Eitzen Aug 22 '13 at 19:57
  • @FlybyNight: I mean if you look at what amy has in the source, it's x^{x^x} indicating that $x$ is raised to the $x^x$ power, but Hagen has a good point about LaTeX allowing {x^x}^x, which would be a bad thing to write. – Jonas Meyer Aug 22 '13 at 20:00
  • The theme here seems to be "don't ask questions that someone else knows the answer to". I was unaware of such a "standard", and that's why I asked. The comments section is intended for people to suggest improvements to a question. My suggestion was greater clarity. – Fly by Night Aug 22 '13 at 20:04
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    @FlybyNight: Has your question not been answered? – Jonas Meyer Aug 22 '13 at 20:14
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    This question has a lively debate on the subject and there is also this wonderful answer. – Dan Brumleve Aug 22 '13 at 20:25
  • @JonasMeyer It was a rhetorical question put in the hope of achieving greater clarity. – Fly by Night Aug 22 '13 at 20:26
  • @DanBrumleve (+1) Thanks... I'll take a look at that. – Fly by Night Aug 22 '13 at 20:27
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    Harvard-MIT tournament, 2012-november, theme, fourth problem. – chubakueno Aug 23 '13 at 00:06
  • "hence the equation has a unique solution" You're making a mistake here. Try substituting in $y = (x+1)^2$ and see if that helps. – user2357112 Aug 22 '13 at 22:31
  • @user2357112: She made no mistake: $x^{x^x}$ really is increasing (obviously so over $x>1$) and so only equals $2012$ for a single value of $x$. The original equation really does have a unique solution for the value of $x^2 + 2x + 3$. –  Aug 22 '13 at 23:26
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    @Hurkyl: There's a unique value of $x^2+2x+3$, but the equation has multiple roots. We're not solving for $x^2+2x+3$. – user2357112 Aug 22 '13 at 23:29

3 Answers3

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The sum is $-2$. Can you see why? Hint: I have not computed the solutions.

Details: As you observed, there is a unique positive $b$ such that $b^{(b^b)}=2012$. Moreover, this $b$ is in the interval $(2,3)$, by the Intermediate Value Theorem, since $2^{(2^2)}$ is too small and $3^{(3^3)}$ is too big.

Note that $x^2+2x+3=(x+1)^2+2$, so $x^2+2x+3$ attains a minimum value of $2$. Thus the equation $x^2+2x+3-b=0$ has two real solutions. The sum of these is the negative of the coefficient of $x$, that is, $-2$.

André Nicolas
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    I will wait a while on the (very small) amount of explanation that would turn the answer into a full solution. I would like to give the OP an opportunity to use the information to discover the solution. – André Nicolas Aug 22 '13 at 20:28
  • the quadratic term is symmetric about x = -1. If there are 2 complex solutions and they are conjugates and their real parts are -1, the imaginary terms cancel when adding and the real parts sum up to -2. But that's not the main reason, I understand that...Just trying to get some clue – imranfat Aug 22 '13 at 20:31
  • @imranfat, more hints: $2^{2^2}<2012$ and V...a – njguliyev Aug 22 '13 at 22:54
  • @imranfat You are right: considering the shape of the function and the symmetry about $-1$, there are exactly two solutions: $r-1$ and $-r-1$. It does not matter what $r$ is, because it cancels out upon addition. – L. F. Aug 22 '13 at 23:05
  • We don't even need to mention the symmetry, though it is nice to do so. Sum of roots. – André Nicolas Aug 22 '13 at 23:08
  • Nice! Indeed, same idea works for bigger towers. If we let $f(x)=x^2+2x+3$, then the sum of the real solutions of $f(x)^{f(x)^{f(x)^{f(x)}}}=2012$ is still $-2$. :) – Prism Aug 23 '13 at 00:23
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    @Prism: We have to be careful with longer towers. The min of $x^2+2x+3$ is $2$. So if the tower has height more than $3$, and we still use $2012$, there is no real root. – André Nicolas Aug 23 '13 at 00:29
  • @AndréNicolas: Ooops you are right :) As there are no real roots, so there is no sum to speak of! But the sum of complex roots would still be $-2$, no? I guess there could a bit of an issue with complex exponentiation... – Prism Aug 23 '13 at 00:44
  • @AndréNicolas: why would the fact that, with a tower of height $4$, $x^2+2x+3=1.9465446382$ alter the sum of the roots? Oops, I see they want the sum of the real roots. – robjohn Aug 23 '13 at 01:21
  • @Prism There would be no problem, because $y=x^2+2x+3$ would be real and positive for any tower, any number $\ge0$. So if they wanted the sum of the roots of a tower of height $h\ge 1$ of $n$th degree polinomials $ax^n+bx^{n-1}+\cdots$, the answer by Viete's formulae would still be $\frac{-b}{a}$ – chubakueno Aug 23 '13 at 01:53
  • @rahul: Yes, thank you for spotting it. – André Nicolas Aug 23 '13 at 09:07
  • And how do we know there is no negative $b$ such that $b^{b^b}=2012$? Indeed, the function $x^{x^x}$ can assume real values for negative $x$. – math_lover Jun 20 '16 at 01:21
  • @JoshuaBenabou: It has been a very long time, so it may take me a while to understand what you mean. Whether powers of $b$ make sense for most negative $b$ can be argued. However, in our case $x^2+2x+3$ is positive for all real $x$, so the issue does not seem to arise. – André Nicolas Jun 20 '16 at 01:30
  • @AndréNicolas: Yes in this case $x^2+2x+3$ is positive, so it's a nonissue, but I was thinking about a harder variant of this problem where we have to look for negative values of $b$ as well. Then it becomes a lot more interesting.It's not clear to me even how to count the number of negative solutions (although I believe there is at least one). I was under the impression that powers of $b$ make sense for all negative $b$, the answer will just be complex most of the time. – math_lover Jun 20 '16 at 03:20
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Hint: Look at Vieta's Formula then compare your equation.

robjohn
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Sketch: $y^{y^y}=2012$ must have only one real solution, call it $a$ (we don't need to compute it). Then we are looking for the roots of $x^2 +2 x+3 - a = (x- x_0)(x-x_1)=x^2 - (x_0+x_1)x +x_0 x_1$ Hence...

leonbloy
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