Does this floor have any affect on the back substitution method while finding the time complexity or we can just ignore it while using back substitution method?

We define S(m) to be T(2m), for every possible value of m. Changing the name of the variable, S(M)=T(2M) for every value of M. Substituting M:=m/2, we get S(m/2)=T(2m/2)

As an example, suppose that T

is the identity function: T(n)=n. We define S(m)=T(2m)=2m. Then S(m/2)=2m/2.

This is what is happening at the last step.Sorry if it is silly but tell me if we are replacing k with log n then won't it happen on the LHS as well and thus instead of T(n) the calculation will be of T(log n)= O (logn log logn)?