(c) ln(x2 - 2x + 1)
= ln((x - 1)2) = 2 ln(x - 1)
Hence, \(\frac{ln(x^2-2x+1)}{ln(x-1)}=\frac{2ln(x-1)}{ln(x-1)}\) = 2 \(\forall\)x >1
Which is a rational number.
(d) \(\frac{ln\sqrt{x-1}}{ln(x-1)}=\cfrac{\frac12 ln(x-1)}{ln(x-1)}\) = 1/2 a rational number