# Finding all solutions of an exponential equation

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

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.

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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