Question

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

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

Answer 1

The correct option is D

$(0,lo{g}_2(4/3\left)\right)\cup (1,\infty )$

Put $2^x=t.$ Then t>0. The given inequality becomes
$\frac{1}{t-1}>\frac{2}{2-t}\Rightarrow \frac{1}{t-1}-\frac{2}{2-t}>0\Rightarrow \frac{2-t-2t+2}{(t-1)(2-t)}>0\Rightarrow \frac{4-3t}{(t-1)(2-t)}>0\Rightarrow (t-1)(t-4/3)(t-2)>0Usingsignchart,weget1<t<4/3ort>2\Rightarrow 1<2^x<4/3or{2}^x>2\Rightarrow 0<x<lo{g}_2\left(4/3\right)orx>1$
Thus, solution set is
$(0,lo{g}_2(4/3\left)\right)\cup (1,\infty )$