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:
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If you raise something to the $0$th power you get $1$, so:
$$\begin{align*}
&x^2 – 2 = 0\\
&(x+\sqrt{2})(x-\sqrt{2}) = 0\\
&x = \pm \sqrt{2}
\end{align*}$$ -
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:
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$x^2-5x-6=0 \Rightarrow x_1 = -1, x_2 = 6$
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$x^2 – 5x + 6 = 0$, $x_1 = 2 \Rightarrow x_2 = 3$ but $x=3$ is not acceptable because $x^2-2 = 7$, odd.
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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
Best Answer
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.