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I came across a really brain-racking problem.

Determine $x$, such that $3^x+3^{-x}=1$.

This is how I tried solving it:

$$3^x+\frac{1}{3^x}=1$$

$$3^{2x}+1=3^x$$

$$3^{2x}-3^x=-1$$

Let $A=3^x$.

$$A^2-A+1=0$$

$$\frac{-1±\sqrt{1^2-4\cdot1\cdot1}}{2\cdot1}=0$$

$$\frac{-1±\sqrt{-3}}{2}=0$$

I end up with

$$\frac{-1±i\sqrt{3}}{2}=0$$

which yields no real solution. And this is not the expected answer.

I'm a 7th grader, by the way. So, I've very limited knowledge on mathematics.

EDIT

I made one interesting observation.

$3^x+3^{-x}$ can be the middle term of a quadratic equation:

$$3^x\cdot\frac{1}{3^x}=1$$

$$3^x+3^{-x}=1$$

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Just building upon previous comments, it doesn't have as you pointed out a Real result but the Imaginary solution can be analytically found as:

$3^x = \dfrac{1\pm\sqrt{3} i}{2}$

Reexpressing rhs in polar notation:

$3^x = e^{i \dfrac{\pi}{3}}$

And changing lhs basis to $e$

$3^x = e^{\ln{3^x}} = e^{x\ln{3}} $

Then:

$\boxed{x = i \dfrac{\pi}{3\ln{(3)}}}$

Note: this is the principal value solution. Due to the periodicity of the function, any $x = i \dfrac{\pi}{3\ln{(3)}} + i \dfrac{2\pi n}{\ln{3}}$, for $n\in \mathbb{Z}$ will also be a solution. Also the 2nd quadrant values need to be considered $x = - i \dfrac{\pi}{3\ln{(3)}} + i \dfrac{2\pi n}{\ln{3}}$

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  • $\begingroup$ Of course, if one embarks on finding complex solutions, there are vastly more solutions than this one. $\endgroup$ – Did Feb 4 '17 at 22:15
  • $\begingroup$ Yes, there are countable infinite solutions due to $2\pi$ periodicity, everyone can seize this detail, otherwise write a post about it, for the moment it seems that the development helped some people to understand how to tackle the problem $\endgroup$ – Arnold Frenzy Feb 5 '17 at 0:17
  • $\begingroup$ Then correct your post (since this is how the site works). $\endgroup$ – Did Feb 5 '17 at 10:18
  • $\begingroup$ The edit is wrong, these are not the solutions. (Unsurprisingly, the OP's extreme agressiveness in now deleted comments correlates well with somewhat shaky mathematical understandings.) $\endgroup$ – Did Feb 5 '17 at 12:42
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    $\begingroup$ you should edit "we" for "I", and which insults? You have thin skin, I was just describing, but nevermind, take my apologies if you think that calling you pedantic was disrespectful and not a description of your behavior and let's move on $\endgroup$ – Arnold Frenzy Feb 5 '17 at 23:05
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Hints (why it's impossible in reals):

  • $3^{x} \gt 0$ for any real $\forall x \in \mathbb{R}$

  • $a + \cfrac{1}{a} \ge 2$ for any positive real $\forall a \in \mathbb{R}^+$

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Note that we have $$ 3^x+3^{-x}=(3^{x/2}-3^{-x/2})^2+2 $$ Apply this to your equation, and you get $$ (3^{x/2}-3^{-x/2})^2=-1 $$ Which means that $3^{x/2}-3^{-x/2}$ is imaginary.

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  • $\begingroup$ You can remove the last line now. $\endgroup$ – Soha Farhin Pine Feb 4 '17 at 8:08
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You have proceeded wrongly. We have $$3^x +3^{-x} =1 $$ $$\Rightarrow 3^{2x} +1 =3^x $$ $$\Rightarrow 3^{2x} -3^x +1=0$$ giving us $$3^x = \frac {1\pm \sqrt {3}i}{2}$$

Notice that the LHS is always real but we have an imaginary part on the RHS. This is enough evidence to suggest that the above equation has no solutions. Hope it helps.

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Note that the function $f(x)=3^x+3^{-x}$ has first derivative $f'(x)=3^x-3^{-x}$ and second derivative $f''(x)=3^x+3^{-x}$. Since $f''(x)>0$ always, this is a convex function that attains a global minimum wherever $f'(x)=0$, so wherever $3^x=3^{-x}$. This occurs at $x=0$, in which the function takes value $f(0)=3^0+3^0=1+1=2$. Therefore, $f(x)\geq2$ for all $x\in\mathbb{R}$ and there does not exist any $x$ such that $3^x+3^{-x}=1$.

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