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Find the sum of all solutions of $(x^2+5x+5)^{(x-3)(x-7)}=1$

An obvious approach is to consider the case where the exponent is $0$ which yields the solutions $3,7$ and then the case where the base is $1$ which yields the solutions $-1,-4$ and then finally to consider the case where base is $-1$ and the exponent is even,which yields the solution $-3$. But the thing is,how do we know these are the only solutions.I don't see any reason these should be the only solutions.Why can't there be complex solutions,or perhaps even more real solutions.Graphing might give us insight,but I don't have any good graphing software at the moment.

I have also thought about using Vieta's formulas,but that seems to complicate things.

Any help,hint or prod in the right direction will be appreciated.Also,is there any way we can know about how the graph of the function will behave without actually graphing the function?

rah4927
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4 Answers4

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I will explain how to find the real solutions. You already found all of them. I don't think there are complex solutions, but I had not managed to prove it.

Let $f(x) = (x^2 + 5x + 5)^{(x - 3)(x - 7)}$, and $\alpha = -\frac{5 + \sqrt 5}2,~\beta = \frac{-5 + \sqrt 5}2$, that is, the solutions to $x^2 + 5x + 5$. The domain of the function $f$ is $$\operatorname{dom} f = (-\infty,~\alpha)\cup\{x \in (\alpha,~\beta) : (x-3)(x-7) \in \mathbb Z\}\cup(\beta,~+\infty).$$ Use the fact that $$x = \begin{cases} \quad\exp(\ln(x)) & x > 0\\ -\exp(\ln(-x)) & x < 0 \end{cases}$$

In your case, we have: $$f(x) = \begin{cases} \quad\exp((x - 3)(x - 7)\ln(x^2 + 5x + 5)) = \kappa(x) & x \in (-\infty,~\alpha) \cup (\beta,~+\infty)\\ -\exp((x - 3)(x - 7)\ln(-x^2 - 5x - 5)) = \chi(x) & x \in (\alpha,~\beta) \end{cases}.$$ You can verify by substitution that $\alpha$ and $\beta$ are not solutions to the original equation.

Therefore we have to solve $$\begin{cases} x \in (-\infty,~\alpha) \cup (\beta,~+\infty)\\ \kappa(x) = 1 \end{cases} \quad\lor\quad \begin{cases}x \in (\alpha,~\beta)\\ \chi(x) = 1 \end{cases}$$

The second system has no solution in the real numbers. As for the first one, $$\begin{cases} x \in (-\infty,~\alpha) \cup (\beta,~+\infty)\\ x = 3 \lor x = 7 \lor x \in \{-4, -3, -2, -1\} \end{cases}$$

Which gives $$x \in \{-4, -1, 3, 7\}.$$

We didn't however consider the case in which the base is negative and the exponent is a number. We are interested in the case in which the base is equal to $-1$ and the exponent is even. This gives, as you mentioned in the question, $x = -3$.

Therefore all the real solutions are $$x \in \{-4, -3, -1, 3, 7\}$$ and their sum is $2$.

rubik
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If $a^x=1$ where $a$ is real

either $x=0$

or $a=1$

or $a=-1$ and $x$ is even

lab bhattacharjee
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Hint:

Use logarithm

$(x-3)(x-7)\log{(x^2+5x+5)}=0$

Logarithm definition can extend to negative numbers too.

Arashium
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    That would work if the thing inside the logarithm is real and positive – rah4927 Feb 21 '15 at 09:14
  • @rah4927, No $log(-x)=log(x)+\pi i$ – Arashium Feb 21 '15 at 09:16
  • This only works if $x^2+5x+5$ is positive. The definition of non-integer powers of negative numbers is itself an issue - even taking the square root of $-1$ you need to make a choice. – Mark Bennet Feb 21 '15 at 09:16
  • Arashium - why not $+2n\pi i$ for arbitrary $n$? – Mark Bennet Feb 21 '15 at 09:17
  • @MarkBennet Seems you are right $\log(-1)=(2n-1 )\pi i$ – Arashium Feb 21 '15 at 09:22
  • No difference. we Just need to find out if each term can be zero or not. – Arashium Feb 21 '15 at 09:24
  • Also,is there any possibility of x being complex? – rah4927 Feb 21 '15 at 09:28
  • @rah4927, one of the 3 terms must be zero. x=3 or x=7 are two answers. the rest of answers must cause inside the logarithm to be 1. do you think there is any solution out of equation $(x^2+5x+5)==0$?? in another word does $\log(y)=0$ has any solution except for y? (including complex ones) – Arashium Feb 21 '15 at 09:35
  • @rah4927 seems $\log(y)=0$ has no other answer except for $y=1$ and $x$ cannot be complex: http://www.wolframalpha.com/input/?i=Solve%5BLog%28x%29%3D%3D0%2Cx%2CComplexes%5D – Arashium Feb 21 '15 at 09:41
  • @Arashium,thanks.I guess I will know more when I start reading in detail about complex numbers. – rah4927 Feb 21 '15 at 09:44
  • How about $\log1=2k\pi i$, so the right-hand side is not zero? – Empy2 Feb 21 '15 at 13:32
  • @Michael good point. We can have non-zero power for non-unit base while having output 1. such as e$^{2 \pi i}$. If we consider $x=a+b i$ then we need to find an integer $k$ matching $$k=\frac{[(a^2-b^2-10a+21)+(2 ab-10b)i]\log[(a^2-b^2+5 a +5)+(2ab+5b)i]}{2 \pi i}$$ But I am stuck in finding $a$ and $b$. At least we know $a$ and $b$ must be real. – Arashium Feb 21 '15 at 14:01
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Taking "$\ln$" on both sides we get,

$(x-3)(x-7)\ln(x^2+5x+5)=\ln 1$ [Assuming the term inside $\ln$ is positive]

$\implies (x-3)(x-7)\ln(x^2+5x+5)= 0$

Now there may arise 3 cases....

The two are easy to find... $x=3,7$

Now what can we do with the complex one,

$\ln(x^2+5x+5)=0$

$\implies x^2+5x +5 =1$

$\implies x^2+5x +4 =0 $

Oops!!!! We get the same result....

rubik
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FAM
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    This is the exact same answer as Arashium's. – rubik Feb 21 '15 at 09:18
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    You are forgetting the cases where the thing inside the log is negative. – rah4927 Feb 21 '15 at 09:19
  • If $x^2+5x+5$ is negative, then we can always write that.. .. $-(x^2+5x+5)$ and then we can take the terms in the left hand side to the right hand side.... Which would make the the power term positive...... – FAM Feb 21 '15 at 09:24
  • @rubik,not really.This answer does not consider certain aspects of the problem(see my previous comment) – rah4927 Feb 21 '15 at 09:32
  • If we assume that the x is either negative or positive, I think the term $x^2+5*x+5$ will always be positive... – FAM Feb 21 '15 at 09:33
  • @AnishDarnal: $x^2 + 5x + 5$ is negative when $$x \in \left(-\frac{5 + \sqrt 5}2,~\frac{-5 + \sqrt 5}2\right)$$ – rubik Feb 21 '15 at 16:46