solving logarithmic equation $\left(\frac{\log x}{2}\right)^{(\log^2x + \log x^2 - 2)} = \log \sqrt{x}$ Solve for $x$

$$\left(\frac{\log x}{2}\right)^{(\log^2x + \log x^2 - 2)} = \log \sqrt{x} $$
Assume that $\log$ is the natural logarithm and $x &gt; 1$ (Otherwise, it would not make sense the power in LHS). Since $\log \sqrt{x} = \dfrac{\log x}{2}$, we can simplify the equation as $$\left(\frac{\log x}{2}\right)^{(\log^2x + \log x^2 - 2)} = \log \sqrt{x}  \implies \left(\frac{\log x}{2}\right)^{(\log^2x + \log x^2 - 3)} = 1$$ 
$$\implies \left(\log^2x + 2\log x - 3\right)\log\left(\dfrac{\log x}{2}\right) = 0.$$

Thus, $$\log\left(\dfrac{\log x}{2}\right) = 0 \implies x = e^{2}.$$ Or,
$$\log^2x + 2\log x - 3 = 0 \implies \log x = -1\pm2  \implies x = e ~~\text{and}~~x = e^{-3}.$$
$\frac{\log x}{2}=\log{\sqrt{x}}$

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