Where/how did a $\log(n)$ factor disappear from well-known algorithms?
You're not wrong, you're just using a different cost model.

Typically there are two:


  
  Uniform cost model - assigns a constant cost to every machine operation regardless of size of numbers.
  Logarithmic cost model - assigns a cost proportional to the number of bits involved for each machine operation.
  


Under uniform cost, a binary search takes constant operations to find the next index to search at, we get a recurrence relations like so:
$$\begin{align}
T(1) &amp; = O(1)\\[0.5em]
T(n) &amp; = T\left( \left\lfloor \frac{n}{2} \right\rfloor \right) + O(1)\\[0.5em]
&amp; = \Theta(\log n)
\end{align}$$

Under logarithmic cost, we can assume it takes $f(n)$ operations to find the next midpoint and we get a recurrence like so: 
$$\begin{align}
T(1) &amp; = O(1)\\[0.5em]
T(n) &amp; = T\left( \left\lfloor \frac{n}{2} \right\rfloor \right) + f(n)\\[0.5em]
\end{align}$$
For binary search, typically $f(n)$ will be no more than computing $\lfloor \frac{l + r}{2} \rfloor$ which would take $\log_2 n$ operations for the addition, and to divide by $2$ we could do a bit shift taking $\log_2 n$ operations (see here). Which brings us to:
$$\begin{align}
T(n) &amp; = T\left( \left\lfloor \frac{n}{2} \right\rfloor \right) + \Theta(\log n)\\[0.5em]
&amp; = \Theta(\log^2 n)
\end{align}$$
These logarithmic costs are just for addition and shifting though. If you were to do other $n$-bit operations then $f(n)$ could change. For example, $n$-bit multiplication and exponentiation could be larger than $\Omega(n)$ and you would have to adjust you complexity accordingly under logarithmic cost.