Question

The number of solution(s) of the equation ${\log }_3\left(x^2-3x-3\right)=\sqrt{{\log }_{\frac{1}{2}}\left(1+\sqrt{x^2-1}\right)}$ is (are)

• A. 0
• B. 1
• C. 2
• D. 4

Answer 1

Answer: (B)

${\log }_3\left(x^2-3x-3\right)=\sqrt{{\log }_{\frac{1}{2}}\left(1+\sqrt{x^2-1}\right)}$
$1+\sqrt{x^2-1}\ge 1\Rightarrow {\log }_{\frac{1}{2}}\left(1+\sqrt{x^2-1}\right)\le 0$
$\sqrt{{\log }_{\frac{1}{2}}\left(1+\sqrt{x^2-1}\right)}$ will be defined only when $x=\pm 1$
Now, at $x=1,x^2-3x-3$ is negative $\Rightarrow$ LHS is not defined and at $x=-1,x^2-3x-3=1+3-3=1\Rightarrow$ LHS is zero
$\therefore$ at $x=-1,LHS=RHS=0$
Hence, $x=-1$ is only the solution