Question 153135
Note that {{{16/sqrt(x-1)=16(x-1)^(-1/2)}}}
So
{{{log(16/sqrt(x-1))}}}
={{{log(16(x-1)^(-1/2))}}}
={{{log(16)+log((x-1)^(-1/2))}}}
={{{log(16)-(1/2)log((x-1))}}}