Question

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

• A. $(1,\infty )$
• B. $(0,{\log }_2(4/3\left)\right)$
• C. $(-1,\infty )$
• D. $(0,1)$

Answer 1

The correct option is B $(0,{\log }_2(4/3\left)\right)$

$\frac{1}{2^x-1}>\frac{1}{1-\frac{2^x}{2}}$

$\frac{1}{2^x-1}-\frac{1}{1-\frac{2^x}{2}}>0$

$let{2}^x=t$

$\frac{1}{t-1}-\frac{2}{2-t}>0$

$\frac{1}{t-1}+\frac{2}{t-2}>0$

$\frac{t-2+2t-2}{(t-1)(t-2)}>0$

$\frac{(t-\frac{4}{3})}{(t-1)(t-2)}>0$

$t\epsilon (1,\frac{4}{3})\bigcup (2,\infty )$

$2^x\epsilon (1,\frac{4}{3})\bigcup (2,\infty )$

$x{\log }_{2}2\epsilon (0,{\log }_2\frac{4}{3})$

$x\epsilon (0,{\log }_2\frac{4}{3})$