If {log _{0.04}}(x - 1) ge {log _{0.2}}(x - 1 | vedclass

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

Question

(c) ${\log _{0.04}}(x - 1) \ge {\log _{0.2}}(x - 1)$ &hellip;..$(i)$

For log to be defined $x - 1 > 0 \Rightarrow x > 1$

From $(i),$ ${\lo

Vedclass · Accepted Answer

$\left[ {2, + \,\infty } \right)$

Vedclass · Answer

(c) ${\log _{0.04}}(x - 1) \ge {\log _{0.2}}(x - 1)$ &hellip;..$(i)$

For log to be defined $x - 1 > 0 \Rightarrow x > 1$

From $(i),$ ${\log _{{{(0.2)}^2}}}(x - 1) \ge {\log _{0.2}}(x - 1)$

$ \Rightarrow $${1 \over 2}{\log _{0.2}}(x - 1) \ge {\log _{0.1}}(x - 1)$

$ \Rightarrow $$\sqrt {x - 1} \le (x - 1)$

$ \Rightarrow $$\sqrt {x - 1} (1 - \sqrt {x - 1} ) \le 0$ $ \Rightarrow $ $1 - \sqrt {x - 1} \le 0$

$ \Rightarrow $$\sqrt {x - 1} \ge 1$ $ \Rightarrow $ $x \ge 2$,

$	herefore \,\,x \in [2,\,\infty )$ .

If ${\log _{0.04}}(x - 1) \ge {\log _{0.2}}(x - 1)$ then $x$ belongs to the interval

Similar Questions

The value of $6+\log _{\frac{3}{2}}\left(\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \ldots}}}\right)$ is

If ${\log _{\tan {{30}^ \circ }}}\left( {\frac{{2{{\left| z \right|}^2} + 2\left| z \right| - 3}}{{\left| z \right| + 1}}} \right)\, < \, - 2$ then

The solution of the equation ${\log _7}{\log _5}$ $(\sqrt {{x^2} + 5 + x} ) = 0$

If ${\log _{0.3}}(x - 1) < {\log _{0.09}}(x - 1),$ then $x$ lies in the interval

If $x = {\log _5}(1000)$ and $y = {\log _7}(2058)$ then