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

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

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

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(C) Given the inequality: ${\log _{0.04}}(x - 1) \ge {\log _{0.2}}(x - 1)$.First,for the logarithm to be defined,we must have $x - 1 > 0$,which implies $x > 1$.We can rewrite the base $0.04$ as $(0.2)^2$. Thus,${\log _{(0.2)^2}}(x - 1) \ge {\log _{0.2}}(x - 1)$.Using the property ${\log _{a^n}}b = \frac{1}{n}{\log _a}b$,we get $\frac{1}{2}{\log _{0.2}}(x - 1) \ge {\log _{0.2}}(x - 1)$.Rearranging the terms: $0 \ge {\log _{0.2}}(x - 1) - \frac{1}{2}{\log _{0.2}}(x - 1)$,which simplifies to $0 \ge \frac{1}{2}{\log _{0.2}}(x - 1)$.This implies ${\log _{0.2}}(x - 1) \le 0$.Since the base $0.2 $x - 1 \ge 1$,which gives $x \ge 2$.Combining this with the domain condition $x > 1$,we get $x \in [2, +\infty)$.

If ${\log _{0.04}}(x - 1) \ge {\log _{0.2}}(x - 1)$,then in which interval does $x$ lie?

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