General Number Field Sieve Big O Clarification
There is a small difference between these two:


$L\approx\exp\left((64/9)^{1/3}\,n^{1/3}\,(\ln(n))^{2/3}\right)$. Here $n$ is a number of bits needed to represent given number $N$.
$O\left( \exp\left((64/9)^{1/3}\,(\log N)^{1/3}\,(\log\log N)^{2/3}\right) \right)$. Here the number $N$ itself is used in notation.


This is true since length of number grows logarithmically of the number itself. Just put $n$ instead of $\log N$ or vice versa and you see that they are equal.

In terms of big O notation constants are uninteresing, so, base of logarithm also is uninteresting. That's why it doesn't make sense whether you interpret number in decimal or binary or any other (except unary) system.

$n=k\,\ln(10)$: here  $\ln(10)$ is constant and we can get rid of it. Also, $k$ depends linearly on $n$, so $k = O(n)$.

$n = \lfloor k\, \log_2(10)\rfloor + 1$: and here both $\log_2(10)$ and $+1$ are removed when written as big O: it still equals $O(n)$.