Solution set of inequality {log _{10}}({x^2} | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(c) ${\log _{10}}({x^2} - 2x - 2) \le 0$&hellip;..$(i)$

For logarithm to be defined,

${x^2} - 2x - 2 > 0$

$ \Rightarrow $ ${(x - 1)^2} &gt

Vedclass · Accepted Answer

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

Vedclass · Answer

(c) ${\log _{10}}({x^2} - 2x - 2) \le 0$…..$(i)$ For logarithm to be defined, ${x^2} - 2x - 2 > 0$ $ \Rightarrow $ ${(x - 1)^2} > 3$ ==> $x - 1 < - \sqrt 3 $ or$x - 1 > \sqrt 3 $ ==> $x < 1 - \sqrt 3 $ or $x > 1 + \sqrt 3 $ i.e., $x < - (\sqrt 3 - 1)$ or $x > (\sqrt 3 + 1)$ Now from $(i),$ ${x^2} - 2x - 2 \le 1$ ==> ${x^2} - 2x - 3 \le 0$ ==> $(x - 3)\,(x + 1) \le 0$ $ \Rightarrow $ $ - 1 \le x \le 3$ $ herefore $ $x \in [ - 1,\, - (\sqrt 3 - 1)\,[\, \cup \,]\,\,\sqrt 3 + 1,\,3]$. i.e., $x \in [ - 1,\,1 - \sqrt 3 )\,\, \cup (1 + \sqrt 3 ,\,3)$.

Solution set of inequality ${\log _{10}}({x^2} - 2x - 2) \le 0$ is

Similar Questions

$\sum\limits_{n = 1}^n {{1 \over {{{\log }_{{2^n}}}(a)}}} = $

If ${1 \over 2} \le {\log _{0.1}}x \le 2$ then

The value of ${81^{(1/{{\log }_5}3)}} + {27^{{{\log }_{_9}}36}} + {3^{4/{{\log }_{_7}}9}}$ is equal to

If $a, b, c$ are distinct positive numbers, each different from $1$, such that $[{\log _b}a{\log _c}a - {\log _a}a] + [{\log _a}b{\log _c}b - {\log _b}b]$ $ + [{\log _a}c{\log _b}c - {\log _c}c] = 0,$ then $abc =$

Let $x, y$ be real numbers such that $x>2 y>0$ and $2 \log (x-2 y)=\log x+\log y$ Then, the possible value(s) of $\frac{x}{y}$