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

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Basic of Logarithms

medium

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

A

$(1, 2)$

B

$(0,1)$

C

($1$, $\infty $)

D

($2$, $\infty )$

Solution

(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 $2$

$\therefore$ $x \in (1,\infty )$.

Standard 11

Mathematics

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