Brain-racking equation

There is no such real number that satisfies the condition.

$\textbf{Why it's impossible in reals?}$

• $3x>0$ for any real $∀x∈R∀$
• $a+\frac{1}{a}≥2$ for any positive real $∀a∈R∀$

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$\textbf{Most people will fall into the trap:}$

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

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

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

$\textbf{Oops! Dead end.}$

Let $A=3^x$ .

$A^2-A+1=0$

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

$\frac{-1±\sqrt{-3}}{2}=0$

$-1±i\sqrt{3}=0$

$\textbf{:-P Got fooled.}$

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True, it doesn't have as a real solution. But the imaginary solution can be analytically found as:

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

Reexpressing R.H.S in polar notation:

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

And changing L.H.S basis to $e$ :

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

Then:

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