<s>[INST] 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:
$$3^x\cdot\frac{1}{3^x}=1$$
$$3^x+3^{-x}=1$$
 [/INST] 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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