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

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Basic of Logarithms

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

A

$(2,\infty )$

B

$(-2, -1)$

C

$(1, 2)$

D

None of these

Solution

(a) ${\log _{0.3}}(x – 1) < {\log _{{{(0.3)}^2}}}(x – 1) = {1 \over 2}{\log _{0.3}}(x – 1)$

$\therefore {1 \over 2}{\log _{0.3}}(x – 1) < 0$

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 $(2,\infty )$.

Standard 11

Mathematics

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