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

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(C) Given the inequality: $\log _{0.04}(x - 1) \ge \log _{0.2}(x - 1)$ $(i)$For the logarithm to be defined,we must have $x - 1 > 0$,which implies $x

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$[2, \infty)$

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(C) Given the inequality: $\log _{0.04}(x - 1) \ge \log _{0.2}(x - 1)$ $(i)$For the logarithm to be defined,we must have $x - 1 > 0$,which implies $x > 1$.Rewrite the base $0.04$ as $(0.2)^2$:$\log _{(0.2)^2}(x - 1) \ge \log _{0.2}(x - 1)$Using the property $\log _{a^n}(b) = \frac{1}{n} \log _a(b)$:$\frac{1}{2} \log _{0.2}(x - 1) \ge \log _{0.2}(x - 1)$Subtract $\frac{1}{2} \log _{0.2}(x - 1)$ from both sides:$0 \ge \frac{1}{2} \log _{0.2}(x - 1)$Since the base $0.2 $x - 1 \ge (0.2)^0$$x - 1 \ge 1$$x \ge 2$Combining with the domain $x > 1$,we get $x \in [2, \infty)$.

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

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