If log_{0.3}(x - 1)

Question

(C) Given the inequality: $\log_{0.3}(x - 1)  0$,which impl

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$(2, \infty)$

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(C) Given the inequality: $\log_{0.3}(x - 1) First,note that for the logarithm to be defined,we must have $x - 1 > 0$,which implies $x > 1$.We can rewrite the base $0.09$ as $(0.3)^2$. Thus,$\log_{0.09}(x - 1) = \frac{\log_{0.3}(x - 1)}{\log_{0.3}(0.09)} = \frac{\log_{0.3}(x - 1)}{2}$.The inequality becomes: $\log_{0.3}(x - 1) Let $y = \log_{0.3}(x - 1)$. Then $y Substituting back: $\log_{0.3}(x - 1) Since the base $0.3$ is between $0$ and $1$,the inequality reverses when we exponentiate: $x - 1 > (0.3)^0$.$x - 1 > 1$,which gives $x > 2$.Therefore,$x \in (2, \infty)$.

If $\log_{0.3}(x - 1) < \log_{0.09}(x - 1)$,then $x \ne 1$ lies in

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