Solving $\sqrt{(\log_2x)^2 - 5|\log_2x| + 6} \le 2\sqrt{5}$

Question

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?

Answer 1

There is nothing wrong with squaring both sides, as long as you keep in mind that you are interested only in those $t$'s such that $t^2-5t+6\geqslant0$, which means that $t\leqslant2$ or that $t\geqslant3$. Since, on the other hand$$t^2-5t+6\leqslant20\iff t\in[-2,7],$$you have\begin{align}\sqrt{t^2-5t+6}\leqslant2\sqrt5&\iff t\in[-2,2]\cup[3,7]\\&\iff\bigl|\log_2(x)\bigr|\in[0,2]\cup[3,7].\end{align}

Answer 2

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.