The set of real values of x for which {log _{ | vedclass

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

Question

(a) ${\log _{0.2}}{{x + 2} \over x} \le 1$&hellip;..$(i)$

For log to be defined, ${{x + 2} \over x} > 0$ $ \Rightarrow $$x > 0$ or $x < -

Vedclass · Accepted Answer

$\left( { - \infty ,\,\, - {5 \over 2}} \right] \cup (0, + \infty )$

Vedclass · Answer

(a) ${\log _{0.2}}{{x + 2} \over x} \le 1$…..$(i)$ For log to be defined, ${{x + 2} \over x} > 0$ $ \Rightarrow $$x > 0$ or $x < - 2$ Now from $(i),$ ${\log _{0.2}}{{x + 2} \over x} \le {\log _{0.2}}0.2$ $ \Rightarrow $${{x + 2} \over x} \ge 0.2$ …..$(ii)$ Case $(i)$ $x > 0$ From $(ii),$ $x + 2 \ge 0.2x$ $ \Rightarrow $ $0.8x \ge - 2$ $ \Rightarrow $$x \ge - {5 \over 2}$. Case $(ii)$ $x < - 2$ From $(ii),$ $x + 2 \le 0.2x$$ \Rightarrow $$0.8x \le - 2$$ \Rightarrow $$x \le - {5 \over 2}$ $ \Rightarrow $$x \in (0,\,\infty )\, \cup \,\left( { - \infty ,\, - {5 \over 2}} ight]$; $ herefore x \in \left( { - \infty ,\, - {5 \over 2}} ight] \cup (0,\,\infty )$.

The set of real values of $x$ for which ${\log _{0.2}}{{x + 2} \over x} \le 1$ is

Similar Questions

If $3^x=4^{x-1}$, then $x=$

$(A)$ $\frac{2 \log _3 2}{2 \log _3 2-1}$ $(B)$ $\frac{2}{2-\log _2 3}$ $(C)$ $\frac{1}{1-\log _4 3}$ $(D)$ $\frac{2 \log _2 3}{2 \log _2 3-1}$

The interval of $x$ in which the inequality ${5^{(1/4)(\log _5^2x)}}\, \geqslant \,5{x^{(1/5)(\log _5^x)}}$

Logarithm of $32\root 5 \of 4 $ to the base $2\sqrt 2 $ is

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

If ${x_n} > {x_{n - 1}} > ... > {x_2} > {x_1} > 1$ then the value of ${\log _{{x_1}}}{\log _{{x_2}}}{\log _{{x_3}}}.....{\log _{{x_n}}}{x_n}^{x_{n - 1}^{{ {\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu \raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1mu}} ^{{x_1}}}}}$ is equal to