Find the product of all the solutions of $\displaystyle\left(\frac{x^2-5x}{6}\right)^{x^2-2}=1$ times the number of solutions.

I don't know how to solve an exponential equation, so I've done as follow:

  1. If you raise something to the $0$th power you get $1$, so:
    $$\begin{align*} &x^2 - 2 = 0\\ &(x+\sqrt{2})(x-\sqrt{2}) = 0\\ &x = \pm \sqrt{2} \end{align*}$$

  2. If the result is $1$ then $\displaystyle\frac{x^2-5x}{6}=\pm1$. When it is equal to $1$ the exponent can be anything, if it is $-1$ it must be even. So:

    • $x^2-5x-6=0 \Rightarrow x_1 = -1, x_2 = 6$

    • $x^2 - 5x + 6 = 0$, $x_1 = 2 \Rightarrow x_2 = 3$ but $x=3$ is not acceptable because $x^2-2 = 7$, odd.

So the solutions are: $S=\{-\sqrt{2}, -1, 2, \sqrt{2}, 6\}$, and the answer to the problem $120$.

Is my work correct? Are there any other methods (simpler, complicated ones)?

EDIT: Wolfram|Alpha does not agree with me:
Wolfram|Alpha results

  • 6
    $\begingroup$ Looks good to me. $\endgroup$ Nov 15, 2011 at 16:27
  • $\begingroup$ Thank you. Why WolframAlpha does not give all the solutions? $\endgroup$
    – rubik
    Nov 15, 2011 at 16:29
  • 1
    $\begingroup$ I just plotted the function, and it looks like those are the only answers, assuming, of course, that $x$ is real. $\endgroup$
    – Phonon
    Nov 15, 2011 at 16:32
  • $\begingroup$ @Phonon: Oh yes I forgot it: $x$ is real! $\endgroup$
    – rubik
    Nov 15, 2011 at 16:34
  • 1
    $\begingroup$ It's fairly standard in elementary algebra and precalculus to restrict the base (whether constant or variable) of an exponentiation to be positive when the exponent is variable. I suspect this convention is behind the Wolfram|Alpha results. $\endgroup$ Nov 15, 2011 at 17:32

2 Answers 2


The easiest way to solve such an equation is taking the logarithm. You will get


and the absolute value is needed to avoid the logarithm will take complex values. Then one has to solve



$$\frac{x^2-5}{6}=\pm 1.$$

This will provide the full set of solutions.

  • $\begingroup$ Thank you, but I haven't studied logarithms yet, so I have some difficulties to understand the step you made (a logarithms property, I guess). Thank you anyway! $\endgroup$
    – rubik
    Dec 7, 2011 at 12:46
  • $\begingroup$ Hi rubik. As you will learn later: this is only valid thanks to the logarithm being an invertible function. It is important to recognize when one uses invertible and non-invertible functions when solving equations, since one may lose solutions or introduce false ones if the function applied is not invertible. $\endgroup$ Feb 11, 2015 at 16:36
  • $\begingroup$ Great method for solving this problem, you deserve more credit for your intuitive thinking. However, since the question asks to find all solutions, I think it is appropriate to extend the logarithm to complex numbers. I also think that the absolute values shouldn't be there because there is no step that called for their need. Instead, you could simply say that the $\log$ of a negative number is something, but when multiplied by $0$, it no longer matters. $\endgroup$ Jan 19, 2016 at 23:04

The fact is that the function $a^x$ is defined only when $a>0$. So firstly you should write $\frac{x^2-5x}{6}\ge0$, so $x\in(-\infty;0)\cup(5;\infty)$. That is why from the solutions you got remain only $x_1=-1$, $x_2=-\sqrt{2}$ and also $x_3=6$. So the set of solutions is $\{-\sqrt2, -1, 6\}$ and the answer to your problem is $18\sqrt2$.

  • $\begingroup$ With $x=2$ you get $(-1)^2=1$ which looks reasonable to me. $\endgroup$
    – Henry
    Nov 15, 2011 at 17:41
  • $\begingroup$ Anyway you have a function of type $a^x$ whose domain is $a>0$. $\endgroup$ Nov 15, 2011 at 17:57
  • $\begingroup$ @Tigran Hakobyan, your reason about a>0 is just why Alpha Wolfram derived only 3 points. But there is a bit difference between equation and function, which I mean $a^n$ still makes sense for a<0, where n is an integer, as Henry just mentioned. However, rubik need to check if LHS makes sense when x equals to some real number. $\endgroup$
    – puresky
    Nov 16, 2011 at 5:59
  • $\begingroup$ @puresky, the equation is given by the function of type $a^x$ which means that we should have $a>0$. In fact, Wolfram Alpha missed the solution $x=6$. Otherwise we could write: $-\sqrt[3](2)=(-2)^{\frac{1}{3}}=(-2)^{\frac{2}{6}}=(4)^{\frac{1}{6}}=\sqrt[6]{4}=\sqrt[3]{2}$, which of course is wrong. $\endgroup$ Nov 16, 2011 at 12:13
  • $\begingroup$ @TigranHakobyan, it seems that you didn't notice I had mentioned that n must be an integer for $a^n$, or a rational if you want to consider complex numbers, when $a<0$. And also I don't think that we need to treat $a^x$ as a function just because there is an x. Dave L. Renfro also didn't say x should be variable. Besides, considering complex, $\sqrt[3]{1}$ is subset of $\sqrt[6]{1}$. $\endgroup$
    – puresky
    Nov 17, 2011 at 3:02

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