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

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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

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