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The problem at hand is, find the solutions of $x$ in the following equation:

$$ (x^2−7x+11)^{x^2−7x+6}=1 $$

My friend who gave me this questions, told me that you can find $6$ solutions without needing to graph the equation.

My approach was this: Use factoring and the fact that $z^0=1$ for $z≠0$ and $1^z=1$ for any $z$.

Factorising the exponent, we have:

$$ x^{2}-7x+6 = (x-1)(x-6) $$

Therefore, by making the exponent = 0, we have possible solutions as $x \in \{1,6\} $

Making the base of the exponent = $1$, we get $$ x^2-7x+10 = 0 $$ $$ (x-2)(x-5)$$

Hence we can say $x \in \{2, 5\} $.

However, I am unable to compute the last two solutions. Could anyone shed some light on how to proceed?

5 Answers5

103

Denote $a=x^2-7x+11.$ The equation becomes $a^{a-5}=1,$ or equivalently* $$a^a=a^5,$$ which has in $\mathbb{R}$ the solutions $a\in \{ {5,1,-1}\}.$ Solving the corresponding quadratic equations we get the solutions $x\in \{1,6,2,5,3,4\}.$

*Note added: $a=0$ is excluded in both equations.

Viera Čerňanová
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    Very elegant way of presenting the problem. In my opinion better than my own answer. I upvote! – Martigan Oct 05 '18 at 08:54
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    hmmmm, what happens if you allow the complex values of $a$ as well? – Carl Witthoft Oct 05 '18 at 16:57
  • It's immediately clear that the solution set given is at least a subset of the full solution set. Is there a simple argument for why there are no other (real) solutions? – Alex Jones Oct 05 '18 at 17:58
  • @AlexanderJ93: that boils down to justifying the fact that $a^a=a^5$ has no other real solutions, which you should be able to do by cases (either $a$ is a positive real number or a negative integer). – Greg Martin Oct 05 '18 at 19:15
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    $0^{-5}$ seems to be $\infty$ or undefined...anyway, a splendid answer. – Lee Oct 06 '18 at 03:40
  • Thank you @Szeto +1, edited. – Viera Čerňanová Oct 06 '18 at 07:56
  • (1 of 2) There are some complex solutions, including ones near $0.23+0.78i$ and $5.8+3.1i$. But we can show there are only $3$ real solutions as follows. Define $f(a)=a^{a-5}-1$. Then $a$ is a solution iff $f(a)=0$. There are no other positive real solutions because $\Re(f)$ is strictly positive for $a\in(0,1)$ and $a\in(5,+\infty)$. Also, there are no other negative real solutions because $\Re(f)<0$ for all $a\in(-\infty,-1)$ while $\Im(f)>0$ for all $a\in(-1,0)$. – Jam Oct 11 '18 at 14:22
  • (2 of 2) For $a$ to be a solution, we need $\Re(f)=\Im(f)=0$. I.e. the blue real curve and red imaginary curve need to meet at the $x$ axis at the same point which only happens in $3$ places. https://i.imgur.com/wefyLdg.jpg – Jam Oct 11 '18 at 14:23
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The only venue you can explore then is to see that $1$ is the power of one other number, that is $-1$.

$(-1)^{2k}=1$, $\forall k\in \mathbb{Z}$

By stating that, you can see that if you have both (you need to fumble around a bit and do a bit of trial and error):

$x^{2}-7x+11=-1$ AND $x^{2}-7x+6=-6$, you would have then $(-1)^{-6}=\frac{1}{(-1)^6}=1$

And of course both equations are in fact the same (otherwise you would not be able to find solutions, that is:

$x^2-7x+12=(x-3)(x-4)=0$ with solutions $(3,4)$.

So $(1,2,3,4,5,6)$ are the six solutions.

Martigan
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Take natural logarithm from both sides: $$\ln (x^2−7x+11)^{x^2−7x+6}=\ln1 \Rightarrow \\ (x^2-7x+6)\cdot \ln |x^2-7x+11|=0 \Rightarrow \\ 1) \ x^2-7x+6=0 \Rightarrow x_{1,2}=1,6; \\ 2) \ \ln |x^2-7x+11|=0 \Rightarrow |x^2-7x+11|=1 \Rightarrow x^2-7x+11=\pm 1 \Rightarrow \\ x_{3,4,5,6}=2,5,3,4.$$ Note: The found solutions satisfy the domain of the equation.

farruhota
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The possibilities are

  • $p^0$: $x^2-7x+6=0\to 1,6$,

  • $1^q$: $x^2-7x+10=0\to 2,5$,

  • $(-1)^q$: $x^2-7x+12=0\to 3,4$, and $q$ is even.

-4

If this is just a casual riddle, then I can agree with the accepted answer. However, if we want to be mathematically strict, I claim that $3$ and $4$ are not solutions of the equation because they lie outside the domain.

Disclaimer: in this post I only consider real exponentiation. It is not my intention to dive into the complex numbers.

What we mean by solving an equation like $f(x) = g(x)$ is finding all $x$ such that both sides make sense and evaluate equal. Hence the first step is always determining the intersection of the domains of $f$ and $g$ - that is, the set of all $x$ such that both sides make sense. Let's consider what that would be in your case.

The left side of your equation naturally decomposes as follows:

$$(x^2-7x+11)^{x^2-7x+6} = p(q_1(x), q_2(x))$$

where

$$\begin{align*} q_1(x) & = x^2-7x+11 \\ q_2(x) & = x^2-7x+6 \\ p(a, b) & = a^b \end{align*}$$

So we have to determine the set of all $(a, b)$ such that $a^b$ makes sense (that is, the domain of exponentiation) and then find the set of all $x$ such that the pair $(q_1(x), q_2(x))$ belongs to this set.

And here is the problem.

There is no uniform way to define both $(-1)^5$ and $3^{\sqrt{2}}$. These are different kinds of exponentiation - the first one is obtained as repeated multiplication, the second one is the result of some limit process, and neither definition works for the other side. So we have a choice: if we allow zero and negative numbers as bases, the exponent must be a non-negative integer, so the domain is $\mathbb{R} \times \mathbb{N}$. If we exclude $0$ as a base, we can use negative exponents, which makes the domain $(\mathbb{R} \setminus \{ 0 \}) \times \mathbb{Z}$. If we go further and exclude negative numbers as bases, we can use limits to pass to real exponents, so the domain becomes $(0, \infty) \times \mathbb{R}$.

One could argue that since the three kinds of exponentiation pairwise agree on the intersections of their domains, we could glue them, i.e. consider the exponentiation on $\mathbb{R} \times \mathbb{N} \cup (\mathbb{R} \setminus \{0\}) \times \mathbb{Z} \cup (0, \infty) \times \mathbb{R}$. But that would be unnatural, useless and - in my opinion - ugly.

Now: which exponentiation does the original equation involve? If the one with natural or integral exponents, then we would have to restrict the domain to those $x$ for which $x^2-7x+6$ is an integer. That doesn't seem right.

Hence we are left with the third, which means we should not consider those $x$ for which $x^2-7x+11$ is negative. This rules out $3$ and $4$ as potential solutions*.

Of course, if we just substitute $x=2$, we get

$$1^{-4} = 1$$

which we know is true and if we substitute $x=3$, we get

$$(-1)^{-6} = 1$$

which we know is equally true, leading to an illusion that both solutions are on equal terms. But that illusion results from using the same notation $a^b$ for two different kinds of exponentiation and it does not stand passing to a strict setting.

*Note: I chose the type of exponentiation that seemed to me to fit in better with the equation. In fact, that choice is an inseparable part of the problem, so it should be disambiguated by the author of the equation (and stated alongside it).

Adayah
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    I get your point of view. Though, $a^{a-5}$ is a correct writing. The trick is in the fact that the base and the exponent are not independent from each other. – Viera Čerňanová Oct 06 '18 at 10:36
  • @user376343 When you write $a^{a-5}$, what kind of exponentiation do you mean (i.e. on what set of pairs $(a, b)$)? – Adayah Oct 06 '18 at 10:41
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    This sounds like madness to me. Everyone knows what $(-1)^5$ and $3^{\sqrt{2}}$ should be, so by definition it cannot be unnatural. Furthermore claiming it is useless is even more wrong, since it clearly is every time $a^b$ comes up before you can or want to pin down what $a$ and $b$ are exactly (for example, this very question). And, well, would you argue that the Riemann Zeta function is unnatural, uesless and ugly? Would you say that about the Kronecker symbol? Would you say that about the Gamma function? Extending functions beyond their original domain is so ubiquitous in math.. – Woett Oct 06 '18 at 10:53
  • @Woett Everyone knows what $(-1)^5$ and $3^{\sqrt{2}}$ should be and everyone would use different definitions to compute those two, so treating them as one thing is unnatural. I strongly disagree with the statement that we write $a^b$ before we know what type of numbers $a, b$ are - and this very question being an example means this question is badly written (if we want to be strict, but as a casual riddle it is ok). – Adayah Oct 06 '18 at 11:27
  • @Woett I wouldn't argue that standard complex functions are ugly - how did that even come to your mind? First, I disclaimed complex numbers in the very beginning of my answer and only spoke of possible definitions of real exponentiation. Second, in both Zeta and Gamma functions we need $a^b$ where $(a, b) \in (0, \infty) \times \mathbb{C}$ for which exponentiation works well, but extending to negative bases would cause problems again. The Kronecker symbol only uses exponentiation on $\mathbb{R} \times \mathbb{N}$. So these functions really don't contradict anything in my answer. – Adayah Oct 06 '18 at 11:49
  • @Woett: Note: extending exponentiation to complex numbers is possible on the three very similar domains: $\mathbb{C} \times \mathbb{N}$, $(\mathbb{C} \setminus { 0 }) \times \mathbb{Z}$ and $(0, \infty) \times \mathbb{C}$ and again it is impossible to unify them. So complex numbers don't solve anything. Unless we speak about multi-valued functions, which is yet another topic... – Adayah Oct 06 '18 at 11:56
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    I am not talking about exponentiation or complex numbers per se. I am talking about the naturalness and usefulness of extending functions beyond their original domain of definition. In this case, $(-1)^5$ and $3^{\sqrt{2}}$ can only mean one thing, and there is no reason not to let it mean that one thing. – Woett Oct 06 '18 at 12:02
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    @Woett I don't see how extending functions beyond their original domain relates to my answer. My point is that there are three, say, variants of exponentiation on different domains that cannot be unified in a natural way but are denoted by the same symbol $a^b$, which is confusing. It's like I defined an operation $$\mathrm{slg}(x) = \begin{cases} \sqrt{x} & x \geqslant 0 \ \log(-x) & x < 0 \end{cases}$$ and asked you to solve the equation $\mathrm{slg}(x) = x^2 - 5$. Would that be natural? In both cases, the same symbol means different, unrelated things in different parts of the domain. – Adayah Oct 06 '18 at 12:59
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    @Adayah That something has different definitions in different parts of its domain doesn't make it not mathematically meaningful to talk about. – JoshuaZ Oct 07 '18 at 13:22
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    @JoshuaZ Did I write it was meaningless? I just wrote it was ugly and unnatural. @${}$Everybody: do you really think that seaming together three conceptually different notions of exponentiation on the union $\mathbb{R} \times \mathbb{N} \cup (\mathbb{R} \setminus { 0 }) \times \mathbb{Z} \cup (0, \infty) \times \mathbb{R}$ has any advantage (in beauty or usefulness) over having three separate exponentiations? Can anybody give any example in serious mathematics where it is necessary to use this compound exponentiation (in a way which cannot be replaced with one of the three)? – Adayah Oct 07 '18 at 14:10
  • @Adayah , you wrote "However, if we want to be mathematically strict, I claim that 3 and 4 are not solutions of the equation because they lie outside the domain." If you don't mean by that the combined function isn't meaningful, then what did you mean by that? – JoshuaZ Oct 07 '18 at 15:16
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    @JoshuaZ By that, I mean that being mathematically strict requires that we recognize the domain of exponentiation before solving the equation at hand. The standard way to understand the exponentiation - in the context of the equation - is as the function defined on $(0, \infty) \times \mathbb{R}$, hence $3, 4$ are outside the domain of $q_1(x)^{q_2(x)}$. In my answer I argue that if the exponentiation were to be understood differently, it would either have to be one of the two other exponentiations, which have too narrow domains, or the union, which is ugly & useless, hence can't be standard. – Adayah Oct 07 '18 at 16:47
  • I came up with yet another example. In some branches of math, it's useful to define the matrix exponential: $$\exp A = \sum_{n=0}^{\infty} \frac{1}{n!} \cdot A^n$$ for $A \in \mathrm{M}{n \times n}(\mathbb{R})$. If somebody posted an equation $e^x = 1+x$, would you post $x = \begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix}$ as a solution? Why not make $\displaystyle \mathbb{R} \cup \bigcup{n=1}^{\infty} \mathrm{M}_{n \times n}(\mathbb{R})$ the domain of $\exp$? – Adayah Nov 11 '18 at 09:41