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  4. If log1/2(x2+6x+9/2(x+1))<-log2(x+1), then x lies in the interval

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Q. If $log_{1/2}\frac{x^{2}+6x+9}{2\left(x+1\right)}<-log_{2}\left(x+1\right),$ then x lies in the interval

Linear Inequalities

Solution:

The log functions are defined if $\frac{x^{2}+6x+9}{2\left(x+1\right)}>0$ and
$x+1>0$
$ \Rightarrow \frac{\left(x+3\right)^{2}}{2\left(x+1\right)}>0$ and $x+1>0$
$ \Rightarrow x>-1$
Now the inequality is $log_{2^{-1}} \frac{x^{2}+6x+9}{2\left(x+1\right)}< -log_{2}\left(x+1\right)$
$\Rightarrow -log_{2} \frac{x^{2}+6x+9}{2\left(x+1\right)}< -log_{2}\left(x+1\right)$
$\Rightarrow log_{2} \frac{x^{2}+6x+9}{2\left(x+1\right)}> log_{2}\left(x+1\right)$
$\Rightarrow \frac{x^{2}+6x+9}{2\left(x+1\right)} >\left(x+1\right)$
$\Rightarrow \frac{-x^{2}+2x+7}{2\left(x+1\right)}>0$
$\Rightarrow \left(x+1\right)\left(x^{2}-2x-7\right)<0$
$\Rightarrow x^{2}-2x-7<0\,\left[\because x+1>0\right]$
$\Rightarrow -1 - 2\sqrt{2} < x < -1 + 2\sqrt{2}$,
but $x >-1 \Rightarrow -1 < x < -1+2\sqrt{2}$