The solution of this equation is in C

Exercise:

Solve, ${{3}^{{{\ln }^{2}}x}}+{{3}^{\ln x}}+1=0$

Solution: Let $w={{3}^{\ln x}}\Rightarrow {{w}^{2}}+w+1=0\Leftrightarrow {{w}^{2}}+2\frac{1}{2}w+1={{w}^{2}}+2\frac{1}{2}w+\frac{1}{4}-\frac{1}{4}+1=0$

$\Rightarrow {{\left( w+\frac{1}{2} \right)}^{2}}=-1+\frac{1}{4}\Leftrightarrow {{\left( w+\frac{1}{2} \right)}^{2}}=\frac{-3}{4}=\frac{3{{i}^{2}}}{4}\Rightarrow w=\frac{-1\pm i\sqrt{3}}{2}$

But $w=\frac{-1}{2}\pm i\frac{\sqrt{3}}{2}=\left\{ \begin{align}
  & -\frac{1}{2}+i\frac{\sqrt{3}}{2}={{e}^{i\left( \pi -\pi /3 \right)}}={{e}^{i\left( 2\pi /3 \right)}} \\
 &  \\
 & \frac{-1}{2}-i\frac{\sqrt{3}}{2}={{e}^{i\left( \pi +\pi /3 \right)}}={{e}^{i\left( 4\pi /3 \right)}} \\
\end{align} \right.$

So ${{e}^{i\left( 2\pi /3 \right)}}={{3}^{\ln x}}\Rightarrow \frac{2\pi }{3}i=\ln x\ln 3\Leftrightarrow \ln x=\frac{2\pi }{3\ln 3}i\Leftrightarrow x={{e}^{\frac{2\pi i}{\ln 27}}}$

And ${{e}^{i\left( 4\pi /3 \right)}}={{3}^{\ln x}}\Rightarrow \frac{4\pi i}{3}=\ln x\ln 3\Rightarrow \ln x=\frac{4\pi i}{3\ln 3}\Leftrightarrow x={{e}^{\frac{4\pi i}{\ln 27}}}$