Question

If ${\log }_3(x-1)<{\log }_{0.09}(x-1)$ then $x$ lies in the interval:

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

Answer 1

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

If $lo{g}_3(x-1)<lo{g}_{0.09}(x-1)$

$lo{g}_3(x-1)<lo{g}_{0.3}^2(x-1)$

$lo{g}_3(x-1)<\frac{1}{2}lo{g}_{0.3}2(x-1)$

$2lo{g}_3(x-1)<lo{g}_{0.3}2(x-1)$

$(x-1{)}^2>x-1$

$x>2$

Then x lies in the interval $2,\infty$