The number of solution of the equation
$\displaystyle \log_3(a^2-3a-3)=\sqrt{\log_{0.5}(1+\sqrt{a^2-1})}$
Here expression is defined when $a^2-1\geq 0\Longrightarrow (a-1)(a+1)>0$
$a\in(-\infty,-1]\cup [1,\infty)$
Also $\displaystyle a^2-3a-3>0$ and $\displaystyle \log_{3}(a^2-3a-3)\geq 0$
$\displaystyle a^2-3a-3\geq 3^0\Longrightarrow a^2-3a-4\geq 0$
$\displaystyle (a-4)(a+1)\geq 0$
$\displaystyle a\in(-\infty,-1]\cup [4,\infty)$
So we have $a\in(-\infty,-1\)\; \cup [4,\infty)$
Now squaring both side of original equation
$\displaystyle \bigg(\log_{3}(a^2-3a-3)\bigg)^2=\log_{0.5}(1+\sqrt{x^2-1})$
I did not understand how do I solve it after that
Please have a look , thanks
You correctly calculated that $1+\sqrt{a^2-1}$ is defined for $(-\infty,1] \cup [1,\infty)$.
Now notice that $1+\sqrt{a^2-1}> 1$ for $a\ne \pm 1$ and $\log_{.5}$ is decreasing which means $\log_{.5}(1+\sqrt{a^2-1})<\log_{.5}(1)=0$ and $\sqrt{\log_{.5}(1+\sqrt{a^2-1})}$ is not defined.
That means that the expression $\sqrt{\log_{.5}(1+\sqrt{a^2-1})}$ is only defined for $a=\pm1$.