Solving the logarithimic inequality $\log_2\frac{x}{2} + \frac{\log_2x^2}{\log_2\frac{2}{x} } \leq 1$

Question

I tried solving the logarithmic inequality: $$\log_2\frac{x}{2} + \frac{\log_2x^2}{\log_2\frac{2}{x} } \leq 1$$ several times but keeping getting wrong answers.

Answer 1

Let $\log_2 x=A$, then $\log_2 x^2=2\log_2 x=2A$ and $\log_2\frac{2}{x}=\log_22- \log_2x=1-A$. So the given inequality becomes: $$(A-1)+\frac{2A}{1-A} \leq 1.$$ Consequently we get $$\frac{4A-A^2-1}{1-A} \leq 1.$$ Furthermore you get $$\frac{5A-A^2-2}{1-A} \leq 0.$$ Hopefully you can solve from here.

Answer 2

Here are the steps \[ \log_2 \frac{x}{2}+\frac{\log_2 x^{2}}{\log_2 \frac{2}{x}} \le 1 \] \[ \log_2 x -\log_2 2+\frac{2\log_2 x}{\log_2 2-\log_2 x} \le 1 \] \[ \log_2 x -1+\frac{2\log_2 x}{1-\log_2 x} \le 1 \] Let $\alpha= \log_2 x$, then \[ \alpha -1+\frac{2\alpha}{1-\alpha} \le 1 \] \[ \alpha -2+\frac{2\alpha}{1-\alpha} \le 0 \] \[ \frac{(1-\alpha)(\alpha -2)+2\alpha}{1-\alpha} \le 0 \] \[ \frac{5\alpha -\alpha^{2}-2}{1-\alpha} \le 0 \] After solving for $\alpha$, we have the solutions \[ 1<\alpha \le \frac{1}{2}(5+\sqrt{17}) \] \[ \alpha \le \frac{1}{2}(5-\sqrt{17}) \] Which is \[ 1<\log_2 x \le \frac{1}{2}(5+\sqrt{17}) \] \[ \log_2 x \le \frac{1}{2}(5-\sqrt{17}) \] Thus \[ 2< x \le (\sqrt{2})^{5+\sqrt{17}} \] \[ x \le (\sqrt{2})^{5-\sqrt{17}} \]

Answer 3

$$\log_2 x - \log_2 2 + 2\log_2 x - \log_2 2 + \log_2 x \le 1$$ $$4\log_2 x - 2 \le 1$$ $$4\log_2 x \le 3$$ $$\log_2 x \le 3/4 $$ $$x \le 2 \cdot 3/4$$

Answer 4

Put $u=\log_2\left(\dfrac x2 \right)=\log_2x-1$

Note that $\log_2(x^2)=2(u+1)$, and $\log_2\left(\dfrac 2x\right)=-u$.

Hence inequality becomes

$$ \begin{align} u-\dfrac {2(u+1)}u \leq1\\ \dfrac{u^2-3u-2}u \leq0\\ \dfrac{(u-\alpha)(u-\beta)}u \leq 0\\ \end{align}$$

where $\alpha=\dfrac {3+\sqrt{17}}2,\ \beta=\dfrac {3-\sqrt{17}}2\ $ $$ \begin{align} u\leq\beta &,\ \quad 0\leq u\leq\alpha,\\\ \log_2\left(\frac x2\right)\leq \dfrac {3-\sqrt{17}}2 &,\quad \ 0\leq\log_2\left(\frac x2\right)\leq\dfrac{3+\sqrt{17}}2\\ \log_2x\leq \frac{5-\sqrt{17}}2&,\quad \ 1\leq \log_2 x\leq \frac{5+\sqrt{17}}2\\ \end{align}$$ As $x>0$, $$0< x\leq2^{\frac{5-\sqrt{17}}2},\ \quad 2\leq x\leq2^{\frac{5+\sqrt{17}}2}$$