<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] Assume that $\log$ is the natural logarithm and $x > 1$ (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,
$$\log^2x + 2\log x - 3 = 0 \implies \log x = -1\pm2  \implies x = e ~~\text{and}~~x = e^{-3}.$$
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