Solve the equation: $\;$ $3^{\log_3 {\left(\log \sqrt{x}\right)}} - \log x + \log^2 x - 3 = 0$
Given equation: $\;\;$ $3^{\log_3 {\left(\log \sqrt{x}\right)}} - \log x + \log^2 x - 3 = 0$
i.e. $\;$ $\log \sqrt{x} - \log x + \log x \times \log x - 3 = 0$
i.e. $\;$ $\log x^{\frac{1}{2}} - \log x + \log x \times \log x - 3 = 0$
i.e. $\;$ $\log \left(\dfrac{x^{\frac{1}{2}}}{x}\right) + \log x \times \log x - 3 = 0$
i.e. $\;$ $\log x^{\frac{-1}{2}} + \log x \times \log x - 3 = 0$
i.e. $\;$ $\log x \times \log x - \dfrac{1}{2} \log x - 3 = 0$
i.e. $\;$ $2 \times \log x \times \log x - \log x - 6 = 0$
i.e. $\;$ $\log x \left(2 \log x - 1\right) - 6 = 0$ $\;\;\; \cdots \; (1)$
Let $\;$ $\log x = p$ $\;\;\; \cdots \; (2)$
Then equation $(1)$ becomes
$p \left(2p - 1\right) - 6 = 0$
i.e. $\;$ $2p^2 - p - 6 = 0$
i.e. $\;$ $\left(2p + 3\right) \left(p - 2\right) = 0$
i.e. $\;$ $p = \dfrac{-3}{2}$ $\;$ or $\;$ $p = 2$
Substituting the value of $p$ in equation $(2)$ gives
when $\;$ $p = \dfrac{-3}{2}$, $\;$ $\log x = \dfrac{-3}{2}$
But logarithim of a number cannot be negative.
$\therefore \;$ $p = \dfrac{-3}{2}$ $\;$ is not a valid solution.
when $\;$ $p = 2$ $\;$ $\log x = 2$ $\implies$ $x = 10^2 = 100$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{100 \right\}$