Question

The number of values of $x$ satisfying $1+{\log }_5(x^2-9)={\log }_5(x^2+4x+3)$ is

• A. $0$
• B. $1$
• C. $2$
• D. infinitely many

Answer 1

The correct option is B $1$

$1+{\log }_5(x^2-9)={\log }_5(x^2+4x+3)...\left(1\right)$
$\Rightarrow x^2-9>0$ and $x^2+4x+3>0$
$\Rightarrow x\in (-\infty ,-3)\cup (3,\infty )...\left(2\right)$
and $x\in (-\infty ,-3)\cup (-1,\infty )...\left(3\right)$
From $\left(2\right)$ and $\left(3\right)$,
$x\in (-\infty ,-3)\cup (3,\infty )$

From eqn$\left(1\right)$,
${\log }_{5}5+{\log }_5(x^2-9)={\log }_5(x^2+4x+3)$
$\Rightarrow {\log }_5\left(5\right(x^2-9\left)\right)={\log }_5(x^2+4x+3)$
$\Rightarrow 5(x^2-9)=x^2+4x+3\Rightarrow 4x^2-4x-48=0\Rightarrow x^2-x-12=0$
$\Rightarrow x=-3,4$

$\therefore x=4$ is the only solution.