Question

If $\frac{1}{{\log }_4\left(\frac{x+1}{x+2}\right)}>\frac{1}{{\log }_4(x+3)},$ then the sets of values of $x$ that satisfy the expression are

• A. $(-1,\infty )$
• B. $(-3,-2)$
• C. $(-3,1)$
• D. $(-2,-1)$

Answer 1

The correct option is B $(-3,-2)$

$\frac{1}{{\log }_4\left(\frac{x+1}{x+2}\right)}>\frac{1}{{\log }_4(x+3)}\cdots \left(1\right)$
$\log$ is defined when
$x+3>0\Rightarrow x\in (-3,\infty )\cdots \left(i\right)$
And $\frac{x+1}{x+2}>0$
$\Rightarrow x\in (-\infty ,-2)\cup (-1,\infty )\cdots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$
$\therefore x\in (-3,-2)\cup (-1,\infty )\cdots \left(2\right)$

$\text{Case 1:}x\in (-1,\infty )...\left(3\right)$
$\Rightarrow \frac{x+1}{x+2}\in (0,1)\text{and}x+3\in (2,\infty )$
$\Rightarrow {\log }_4\left(\frac{x+1}{x+2}\right)<0\text{and}{\log }_4(x+3)>0$
which does not satisfy inequation $\left(1\right)$

$\text{Case 2:}x\in (-3,-2)...\left(4\right)$
$\Rightarrow \frac{x+1}{x+2}\in (2,\infty )\text{and}-x+3\in (0,1)$
$\Rightarrow {\log }_4\left(\frac{x+1}{x+2}\right)>0\text{and}{\log }_4(x+3)<0$
$\Rightarrow {\log }_4\left(\frac{x+1}{x+2}\right)>{\log }_4(x+3)$
$\Rightarrow \frac{x+1}{x+2}>x+3$
$\Rightarrow \frac{x+1}{x+2}-x+3>0$
$\Rightarrow \frac{-(x^2+4x+5)}{x+2}>0$
$\Rightarrow \frac{(x+2{)}^2+1}{x+2}<0$
$\Rightarrow x<-2...\left(5\right)$

Now, from $\left(2\right),\left(4\right)$ and $\left(5\right)$
$x\in (-3,-2)$