Question

How many real values of $x$ exist such that ${\log }_{(x+4)}(x^2-1)={\log }_{x+4}(5-x)$ holds good ?

• A. $0$
• B. $1$
• C. $2$
• D. Infinitely Many

Answer 1

The correct option is B $1$

${\log }_{(x+4)}(x^2-1)={\log }_{x+4}(5-x)...\left(1\right)$
$\Rightarrow x^2-1>0,5-x>0,x+4>0$ and $x+4\ne 1$
$\Rightarrow x\in (-\infty ,-1)\cup (1,\infty ),x<5,x>-4,x\ne -3\Rightarrow x\in (-4,-3)\cup (-3,-1)\cup (1,5)$


From eqn $\left(1\right),$
$x^2-1=5-x$
$\Rightarrow x^2+x-6=0\Rightarrow (x+3)(x-2)=0$
$\Rightarrow x=-3,x=2$
Since, $x\ne -3$
$\therefore x=2$ is the only solution.