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Basic of Logarithms
easy
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
