Question

If ${\log }_{\frac{1}{\sqrt{2}}}(x-1)>2,$ then $x$ lies in the interval

• A. $(\frac{3}{2},\infty )$
• B. $(-\infty ,\frac{3}{2})$
• C. $(\frac{3}{2},1)$
• D. None of these

Answer 1

The correct option is B $(-\infty ,\frac{3}{2})$

Given: ${\log }_{\frac{1}{\sqrt{2}}}(x-1)>2$

Since base $\frac{1}{\sqrt{2}}<1$
Hence, the sign of inequality is reversed.
$\therefore x-1<{\left(\frac{1}{\sqrt{2}}\right)}^2\Rightarrow x<\frac{3}{2}$
$\therefore x\in (-\infty ,\frac{3}{2})$