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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

jacky
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  • It looks like you meant to write "So we have $a \in (-\infty, -1] \cup [4, \infty)$. Formatting tip: You can obtain $\implies$ by typing $\implies$. – N. F. Taussig Jun 11 '23 at 12:09

1 Answers1

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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 $a=\pm1$.

jjagmath
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