Question

The complete set of values of $x$ that satisfies the expression $\frac{1}{{\log }_4\left(\frac{x+1}{x+2}\right)}<\frac{1}{{\log }_4(x+3)}$ is

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

Answer 1

The correct option is A $(-1,\infty )$

$\log$ is defined when
$x+3>0\Rightarrow x>-3\Rightarrow x\in (-3,\infty )$
And $\frac{(x+1)}{(x+2)}>0\Rightarrow (x+1)(x+2)>0\Rightarrow x\in (-\infty ,-2)\cup (-1,\infty )$
$\therefore x\in (-3,2)\cup (-1,\infty )$ ...(1)
Now, ${\log }_4(x+3)<{\log }_4\frac{(x+1)}{(x+2)}\Rightarrow (x+3)<\frac{(x+1)}{(x+2)}$
When $x+2>0\Rightarrow x>-2\Rightarrow x\in (-2,\infty )$ ...(2)
$(x+3)(x+2)>(x+1)\Rightarrow x^2+4x+5>0$ ( this is always true)
Now from (1) and (2)
$x\in (-1,\infty )$