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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

  • 1
    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.2011-11-15

2 Answers 2

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

$(x^2-2)\log\left(\left|\frac{x^2-5}{6}\right|\right)=0$

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

$x^2-2=0$

and

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

This will provide the full set of solutions.

  • 0
    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.2016-01-19
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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 $18\sqrt2$.

  • 0
    @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 $\sqrt[6]{1}$.2011-11-17