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

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Question

(c) ${\log _{0.3}}(x - 1) < {\log _{0.09}}(x - 1)$

$1 < \frac{{{{\log }_{0.09}}(x - 1)}}{{{{\log }_{0.3}}(x - 1)}}$

==> $1 < {\log _

Vedclass · Accepted Answer

($1$, $\infty $)

Vedclass · Answer

(c) ${\log _{0.3}}(x - 1) < {\log _{0.09}}(x - 1)$ $1 < \frac{{{{\log }_{0.09}}(x - 1)}}{{{{\log }_{0.3}}(x - 1)}}$ ==> $1 < {\log _{0.3}}(0.09)$ ==> $1 < {\log _{0.3}}{(0.3)^2}$ ==> $1 < 2$ Which of true therefore it is true for every positive value of $2$ $ herefore$ $x \in (1,\infty )$.

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

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