What is $x$, if $3^x+3^{-x}=1$? 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:
$$3^x\cdot\frac{1}{3^x}=1$$
$$3^x+3^{-x}=1$$
 A: 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 $\forall a \in \mathbb{R}^+$
A: 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.
A: 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. 
A: 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 $x = - i  \dfrac{\pi}{3\ln{(3)}} + i \dfrac{2\pi n}{\ln{3}}$
A: 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 $3^x+3^{-x}=1$.
