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:
$$3^x\cdot\frac{1}{3^x}=1$$
$$3^x+3^{-x}=1$$

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 $\forall a \in \mathbb{R}^+$
