If {log _{0.3}}(x - 1)

Question

(A) Given the inequality: ${\log _{0.3}}(x - 1)  0$,so $x > 1$.Rewrite the base $

Vedclass · Accepted Answer

$(2, \infty)$

Vedclass · Answer

(A) Given the inequality: ${\log _{0.3}}(x - 1) First,note that the domain requires $x - 1 > 0$,so $x > 1$.Rewrite the base $0.09$ as $(0.3)^2$: ${\log _{0.09}}(x - 1) = \frac{1}{2}{\log _{0.3}}(x - 1)$.The inequality becomes: ${\log _{0.3}}(x - 1) Subtracting $\frac{1}{2}{\log _{0.3}}(x - 1)$ from both sides: $\frac{1}{2}{\log _{0.3}}(x - 1) This implies ${\log _{0.3}}(x - 1) Since the base $0.3  (0.3)^0$,which means $x - 1 > 1$.Thus,$x > 2$.Therefore,$x$ lies in the interval $(2, \infty)$.

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

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