The solution set of the inequality {log _{10} | vedclass

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Question

(C) Given inequality: ${\log _{10}}({x^2} - 2x - 2) \le 0$. Since the base $10 > 1$,the inequality remains the same when removing the log: ${x^2} - 2x

Vedclass · Accepted Answer

$[ - 1, 1 - \sqrt 3 ) \cup (1 + \sqrt 3, 3]$

Vedclass · Answer

(C) Given inequality: ${\log _{10}}({x^2} - 2x - 2) \le 0$. Since the base $10 > 1$,the inequality remains the same when removing the log: ${x^2} - 2x - 2 \le {10^0} \implies {x^2} - 2x - 2 \le 1 \implies {x^2} - 2x - 3 \le 0$. Factoring the quadratic: $(x - 3)(x + 1) \le 0$,which gives $x \in [-1, 3]$. Also,the argument of the log must be positive: ${x^2} - 2x - 2 > 0$. Roots of ${x^2} - 2x - 2 = 0$ are $x = \frac{2 \pm \sqrt{4 - 4(1)(-2)}}{2} = 1 \pm \sqrt{3}$. So,${x^2} - 2x - 2 > 0$ for $x \in (-\infty, 1 - \sqrt{3}) \cup (1 + \sqrt{3}, \infty)$. Taking the intersection of $x \in [-1, 3]$ and $x \in (-\infty, 1 - \sqrt{3}) \cup (1 + \sqrt{3}, \infty)$,we get: $x \in [-1, 1 - \sqrt{3}) \cup (1 + \sqrt{3}, 3]$.

The solution set of the inequality ${\log _{10}}({x^2} - 2x - 2) \le 0$ is:

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