Question

If ${\log }_4(x^2+x)-{\log }_4(x+1)=2$, then the value of $x$ is equal to

• A. $1$
• B. $2$
• C. $4$
• D. $16$

Answer 1

Answer: (D)

$16$

Domain of the function:
$x^2+x>0$ and $x+1>0$
$\Rightarrow x(x+1)>0\Rightarrow x\in (-\infty ,-1)\cup (0,\infty )$
And $x+1>0\Rightarrow x\in (-1,\infty )$
Taking intersection, we have
$\Rightarrow x\in (0,\infty )$
Now,
${\log }_4(x^2+x)-{\log }_4\left(x+1\right)=2$
$\Rightarrow {\log }_4\left(\frac{x^2+x}{x+1}\right)[\because \log a-\log b=\log \frac{a}{b}]=2$
$\Rightarrow \frac{x(x+1)}{x+1}=4^2=16$
$\Rightarrow x=16$
Ans: D