Question

If $lo{g}_{03}(x-1)<lo{g}_{009}(x-1)$, then x lies in the interval

• A. $(2,\infty )$
• B. $(-2,-1)$
• C. $(1,2)$
• D. None

Answer 1

The correct option is A $(2,\infty )$

$lo{g}_{03}(x-1)<lo{g}_{(03{)}^2}(x-1)=\frac{1}{2}lo{g}_{03}(x-1)$
$\therefore \frac{1}{2}lo{g}_{03}(x-1)<0$
or $lo{g}_{03}(x-1)<0=log1$
or $(x-1)>1orx>2$
as base is less than 1, therefore the inequality is reversed, now $x>2\Rightarrow x$ lies in $(2,\infty )$