Solving $\sqrt{(\log_2x)^2 - 5|\log_2x| + 6} \le 2\sqrt{5}$ My attempt:
Given,
$$\sqrt{(\log_2x)^2 - 5|\log_2x| + 6} \le 2\sqrt{5}$$
Taking $t=|\log_2x|$:
$$\sqrt{t^2-5t+6} \le 2\sqrt{5}$$
I know that squaring both sides will lead to extraneous solutions. In fact, I solved the rest of the problem by squaring sides and got $x \in \left[\dfrac{1}{128},128\right]$ whereas the correct answer according to Wolfram Alpha is $x \in \left[\dfrac{1}{128},\dfrac{1}{8}\right] \cup \left[\dfrac{1}{4}, 4\right] \cup [8,128]$.
What is the correct way of solving this problem that avoids extraneous solutions?

This inequality is quite fiddly, but with care we can get to the solution. First, note that any inequality of the form $\sqrt{a}\le b$ is equivalent to $a\le b^2$ and $a\ge0$, meaning
\begin{align}
\sqrt{t^2-5t+6}\le2\sqrt{5} &\iff t^2-5t+6\le20 \quad\text{and}\quad t^2-5t+6\ge0 \\[5pt]
&\iff (t+2)(t-7)\le0\quad\text{and}\quad(t-2)(t-3)\ge0 \\[5pt]
&\iff t\in[-2,7]\quad\text{and}\quad t\not\in(2,3) \\[5pt]
&\iff t\in\left([-2,7]\setminus(2,3)\right)\\[5pt]
&\iff t\in[-2,2]\cup[3,7] \, .
\end{align}
Hence, $-2\le|\log_2(x)|\le 2$ or $3\le|\log_2(x)|\le 7$.
If $x\ge1$, then these inequalities become $-2\le\log_2(x)\le2$ (which has the solution $\frac{1}{4}<x\le4$) and $3\le\log_2(x)\le7$ (which has the solution $8\le x\le 128$). So if $x\ge1$, then $\frac{1}{4}\le x\le 4$ or $8\le x\le 128$. This boils down to $1\le x\le 4$ or $8\le x\le128$.
If $0<x<1$, then the inequalities become $-2\le-\log_2(x)\le 2$ (which has solution $\frac{1}{4}\le x\le2$) and $3\le -\log_2(x) \le 7$ (which has solution $\frac{1}{128}\le x\le\frac{1}{8}$). This boils down to $\frac{1}{128}\le x\le\frac{1}{8}$ or $\frac{1}{4}\le x\le1$.
Combining all of these solutions, we find that $\frac{1}{128}\le x\le\frac{1}{8}$ or $\frac{1}{4}\le x\le 4$ or $8\le x\le 128$.
I will give an example of how I solved one of the above inequalities. The rest of them are left as exercises:
\begin{align}
3\le -\log_2(x)\le 7 &\iff -3 \ge \log_2(x) \ge - 7 \\[5pt]
&\iff -7 \le \log_2(x) \le -3 \\[5pt]
&\iff 2^{-7} \le x \le 2^{-3} \\[5pt]
&\iff \frac{1}{128} \le x \le \frac{1}{8} \, .
\end{align}
Finally, there is a simpler way to solve $-2\le|\log_2(x)|\le 2$ or $3\le|\log_2(x)|\le 7$, but it is not obvious. A number $y$ satisfies these inequalities if and only if $\frac{1}{y}$ satisfies them. So once we have the solutions $1\le x \le 4$ or $8 \le x \le 128$, we can easily find the others. Admittedly, I didn't realise this until I solved this inequality the long-winded way.
