If $\log _{1 / 2} \frac{x^2+6 x+9}{2(x+1)}

Question

If $\log _{1 / 2} \frac{x^2+6 x+9}{2(x+1)}<-\log _2(x+1)$, then $x$ belongs to the interval

Professor PC Thomas Classes · Answer

$ \begin{aligned} & \log _{1 / 2} \frac{x^2+6 x+9}{2(x+1)}<-\log _2(x+1)-\log _2 \frac{x^2+6 x+9}{2(x+1)}<-\log _2(x+1) \ & \quad \operatorname{Domain}_{(x+1)>0} \& \frac{(x+3)^2}{2(x+1)}>0 \Rightarrow x \in(-1, \infty) \ & \quad \log _{1 / 2}\left(\frac{(x+3)^2}{2(x+1)} ight)<-\log _2(x+1) \ & \Rightarrow \frac{(x+3)^2}{2(x+1)}>(x+1) \Rightarrow \frac{((\sqrt{2}+1) x+(3+\sqrt{2}))((1-\sqrt{2}) x+(3-\sqrt{2}))}{2(x+1)}>0 \ & \Rightarrow \frac{(x-(2 \sqrt{2}-1))(x-(2 \sqrt{2}+1)}{(x+1)}<0 \ & \Rightarrow x \in(-1,1+2 \sqrt{2}) \end{aligned} $ \