<s>[INST] solving logarithmic equation $\left(\frac{\log x}{2}\right)^{(\log^2x + \log x^2 - 2)} = \log \sqrt{x}$ Solve for $x$
$$\left(\frac{\log x}{2}\right)^{(\log^2x + \log x^2 - 2)} = \log \sqrt{x} $$
 [/INST] $\frac{\log x}{2}=\log{\sqrt{x}}$
Using this,
$( \log{\sqrt{x}})^{(\log x)^2+2\log x-2}= \log{\sqrt{x}}$
$\implies (\log x)^2+2\log x-2=1$
$\implies (\log x+3)(\log x-1)=0$
$\implies x=e^{-3}$ or $e$
Again, 
$\log{\sqrt{x}}=1$ gives a solution.
$\implies x=e^2$ is also a solution
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