Question

Solve for $x$: ${\log }_{10}(x^2-2x-2)\le 0$

Answer 1

For logarithm to be defined ,
$x^2-2x-2>0$ .....(1)

We will find the roots of $x^2-x-2$ by quadratic formula,

$x=\frac{2\pm \sqrt{4+8}}{2}$

$x=\frac{2\pm 2\sqrt{3}}{2}$

$\Rightarrow x=1\pm \sqrt{3}$

So, inequality (1) can be written as

$[x-(1-\sqrt{3}\left)\right][x-(1+\sqrt{3}\left)\right]>0$

$\Rightarrow x\in (-\infty ,1-\sqrt{3})\cup (1+\sqrt{3},\infty )$ ...(2)

Now, given ${\log }_{10}(x^2-2x-2)\le 0$

$\Rightarrow x^2-2x-2\le {\left(10\right)}^0$
$\Rightarrow x^2-2x-2\le 1$

$\Rightarrow x^2-2x-3\le 0$

$\Rightarrow x^2-3x+x-3\le 0$

$\Rightarrow (x-3)(x+1)\le 0$
$\Rightarrow x\in [-1,3]$ ...(3)

From (2) and (3), we get
$x\in [-1,1-\sqrt{3})\cup (1+\sqrt{3},3]$