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

Question

(a) ${\log _{0.3}}(x - 1) < {\log _{{{(0.3)}^2}}}(x - 1) = {1 \over 2}{\log _{0.3}}(x - 1)$

$	herefore {1 \over 2}{\log _{0.3}}(x - 1) &lt

Vedclass · Accepted Answer

$(2,\infty )$

Vedclass · Answer

(a) ${\log _{0.3}}(x - 1) < {\log _{{{(0.3)}^2}}}(x - 1) = {1 \over 2}{\log _{0.3}}(x - 1)$ $ herefore {1 \over 2}{\log _{0.3}}(x - 1) < 0$ or ${\log _{0.3}}(x - 1) < \,0 = \log 1$ or $(x - 1) > 1$ or $x > 2$ As base is less than $1$, therefore the inequality is reversed, now $x > 2$ $ \Rightarrow $$x$ lies in $(2,\infty )$.

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

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