Question

Find the solution of the inequality ${\log }_{1/2}(x+1)>{\log }_2(2-x)$

• A. $\frac{1+\sqrt{5}}{2}<x<2$
• B. $-1<x<\frac{1-\sqrt{5}}{2}$
• C. $-1<x<\frac{1+\sqrt{5}}{2}$
• D. $\frac{1-\sqrt{5}}{2}<x<2$

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{1+\sqrt{5}}{2}<x<2$
B $-1<x<\frac{1-\sqrt{5}}{2}$

${\log }_{1/2}(x+1)>{\log }_2(2-x)$
$-{\log }_2(x+1)>{\log }_2(2-x)$
$\Rightarrow \frac{1}{x+1}>2-x$
$\Rightarrow x^2-x-1>0$
$\Rightarrow {(x-\frac{1}{2})}^2-\frac{5}{4}>0$
$\Rightarrow [x-(\frac{1}{2}+\frac{\sqrt{5}}{2})][x-(\frac{1}{2}-\frac{\sqrt{5}}{2})]>0$
$\Rightarrow x\in (-\infty ,\frac{1-\sqrt{5}}{2})\cup (\frac{1+\sqrt{5}}{2},\infty )$ ..... $\left(1\right)$
Also, from the given inequality, it follows that
$\frac{1}{x+1}>0$ and $2-x>0$
$\Rightarrow x+1>0$ and $x<2$
$\Rightarrow x>-1$ and $x<2$ ..... $\left(2\right)$
From (1) and (2), it follows that
$-1<x<\frac{1-\sqrt{5}}{2}$ and $\frac{1+\sqrt{5}}{2}<x<2$