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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?

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    $\begingroup$ Hint: If the base is $-1$, and the exponent is an even integer, then . . . $\endgroup$ – quasi Oct 5 '18 at 8:28
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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.

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    $\begingroup$ Very elegant way of presenting the problem. In my opinion better than my own answer. I upvote! $\endgroup$ – Martigan Oct 5 '18 at 8:54
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    $\begingroup$ hmmmm, what happens if you allow the complex values of $a$ as well? $\endgroup$ – Carl Witthoft Oct 5 '18 at 16:57
  • $\begingroup$ 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? $\endgroup$ – AlexanderJ93 Oct 5 '18 at 17:58
  • $\begingroup$ @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). $\endgroup$ – Greg Martin Oct 5 '18 at 19:15
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    $\begingroup$ $0^{-5}$ seems to be $\infty$ or undefined...anyway, a splendid answer. $\endgroup$ – Szeto Oct 6 '18 at 3:40
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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.

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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.

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    $\begingroup$ It should be noted that to go from the first line to the second, you need to use the fact that $x^2-7x+6$ is always even. $\endgroup$ – eyeballfrog Oct 8 '18 at 17:31
  • $\begingroup$ Thanks for the note. It is noted in the note. $\endgroup$ – farruhota Oct 8 '18 at 18:54
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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.

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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).

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    $\begingroup$ 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. $\endgroup$ – user376343 Oct 6 '18 at 10:36
  • $\begingroup$ @user376343 When you write $a^{a-5}$, what kind of exponentiation do you mean (i.e. on what set of pairs $(a, b)$)? $\endgroup$ – Adayah Oct 6 '18 at 10:41
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    $\begingroup$ 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.. $\endgroup$ – Woett Oct 6 '18 at 10:53
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    $\begingroup$ 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. $\endgroup$ – Woett Oct 6 '18 at 12:02
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    $\begingroup$ @Adayah That something has different definitions in different parts of its domain doesn't make it not mathematically meaningful to talk about. $\endgroup$ – JoshuaZ Oct 7 '18 at 13:22

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