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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

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Looks good to me. – Ross Millikan Nov 15 '11 at 16:27
Thank you. Why WolframAlpha does not give all the solutions? – rubik Nov 15 '11 at 16:29
I just plotted the function, and it looks like those are the only answers, assuming, of course, that $x$ is real. – Phonon Nov 15 '11 at 16:32
@Phonon: Oh yes I forgot it: $x$ is real! – rubik Nov 15 '11 at 16:34
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. – Dave L. Renfro Nov 15 '11 at 17:32
up vote 2 down vote accepted

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.

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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! – rubik Dec 7 '11 at 12:46
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. – mathreadler Feb 11 '15 at 16:36
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. – Simple Art Jan 19 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 $18\sqrt2$.

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With $x=2$ you get $(-1)^2=1$ which looks reasonable to me. – Henry Nov 15 '11 at 17:41
Anyway you have a function of type $a^x$ whose domain is $a>0$. – Tigran Hakobyan Nov 15 '11 at 17:57
@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. – puresky Nov 16 '11 at 5:59
@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. – Tigran Hakobyan Nov 16 '11 at 12:13
@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 $\sqrt[6]{1}$. – puresky Nov 17 '11 at 3:02

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