Question

Solution set of the inequality $\frac{1}{2^{x}-1} > \frac{1}{1-2^{x-1}}$ is

• A. $(1, \infty)$
• B. $(0, log_2(4/3))$
• C. $(-1, \infty)$
• D. $(0, log_2(4/3)) \cup (1, \infty)$

Answer 1

Answer: (D)

Put $2^x = t$. Then $t > 0$. Now, given inequality becomes $\frac{1}{t-1} > \frac{2}{2-t} \Rightarrow \frac{1}{t-1}-\frac{2}{2-t} >0 \Rightarrow \frac{2-t-2t+2}{\left(t-1\right)\left(2-t\right)} > 0$
$\Rightarrow \frac{4-3t}{\left(t-1\right)\left(2-t\right)} >0 \Rightarrow \frac{\left(t-\frac{4}{3}\right)}{\left(t-1\right)\left(t-2\right)} >0.$

From sign scheme we get
$1 < t < 4/3$ or $t > 2.$
$\Rightarrow 1 < 2^{x} < 4/3$ or $2^{x} > 2$
$\Rightarrow x \in\left(0, log_{2}\left(4/3\right)\right)\cup\left(1, \infty\right)$