Solve the equation: $\;$ $3 \sqrt{\log_2 x} - \log_2 8x + 1 = 0$
Given equation: $\;\;$ $3 \sqrt{\log_2 x} - \log_2 8x + 1 = 0$
i.e. $\;$ $3 \sqrt{\log_2 x} - \left(\log_2 8 + \log_2 x\right) + 1 = 0$
i.e. $\;$ $3 \sqrt{\log_2 x} - \log_2 2^3 - \log_2 x + 1 = 0$
i.e. $\;$ $3 \sqrt{\log_2 x} - 3 \log_2 2 - \log_2 x + 1 = 0$
i.e. $\;$ $3 \sqrt{\log_2 x} - 3 - \log_2 x + 1 = 0$
i.e. $\;$ $3 \sqrt{\log_2 x} - \log_2 x - 2 = 0$ $\;\;\; \cdots \; (1)$
Let $\;$ $\sqrt{\log_2 x} = p$ $\;\;\; \cdots \; (2)$
Then in view of equation $(2)$, equation $(1)$ becomes
$3p - p^2 - 2 = 0$
i.e. $\;$ $p^2 - 3p + 2 = 0$
i.e. $\;$ $\left(p - 1\right) \left(p - 2\right) = 0$
$\implies$ $p = 1$ $\;$ or $\;$ $p = 2$
Substituting the value of $p$ in equation $(2)$ gives
when $\;$ $p = 1$, $\;$ $\sqrt{\log_2 x} = 1$
i.e. $\;$ $\log_2 x = 1$ $\implies$ $x = 2^1 = 2$
when $\;$ $p = 2$, $\;$ $\sqrt{\log_2 x} = 2$
i.e. $\;$ $\log_2 x = 4$ $\implies$ $x = 2^4 = 16$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{2, \; 16 \right\}$