Question

If $lo{g}_{\frac{1}{\sqrt{2}}}(x-1)>2$ 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 $(-\infty ,\frac{3}{2})$

If $lo{g}_{\frac{1}{\sqrt{2}}}(x-1)>2.1$
$=2lo{g}_{\frac{1}{\sqrt{2}}}\left(\frac{1}{\sqrt{2}}\right)=lo{g}_{\frac{1}{\sqrt{2}}}{\left(\frac{1}{\sqrt{2}}\right)}^2=lo{g}_{\frac{1}{\sqrt{2}}}\left(\frac{1}{2}\right)$
Since base $\frac{1}{\sqrt{2}}$ is less than one and hence the sign of inequality is reversed.
$\therefore x-1<\frac{1}{2}orx<\frac{3}{2}$
$\therefore xϵ(-\infty ,\frac{3}{2})$