Logarithms

Concept:For a logarithmic inequality with base between 0 and 1, the inequality sign reverses when converting to exponential form.Explanation:The base $\frac{1}{\sqrt{2}}$ is between 0 and 1, so the function is decreasing.The argument must be positive: $x-1 > 0 \Rightarrow x > 1$.Given $\log_{\frac{1}{\sqrt{2}}}(x-1) > 2$, we convert:$x-1 < \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$ (inequality reversed).Thus $x < 1 + \frac{1}{2} = \frac{3}{2}$.Combining with $x > 1$, we get $1 < x < \frac{3}{2}$, i.e., $(1, \frac{3}{2})$.This interval is not listed among options A, B, or C.Answer:Option D: none