Is there a way to find the log of very large numbers?
By the laws of logarithms
$$ \log_2(256!) = \sum_{i=1}^{256} \log_2(i)$$
This is easily evaluated (albeit not on all calculators).
In Python:
>>> sum(math.log2(i) for i in range(1,257))
1683.9962872242136
If it's about factorials, you can use Stirling's approximation:
$$\ln(N!) \approx N\ln(N) - N$$
Due to the fact that
$$N! \approx \sqrt{2\pi N}\ N^N\ e^{-N}$$
Error Bound
Writing the "whole" Stirling series as
$$\ln(n!)\approx n\ln(n)−n+\frac{1}{2}\ln(2\pi n)+\frac{1}{12n} −\frac{1}{360n^3}+\frac{1}{1260n^5}+\ldots $$
it is known that the error in truncating the series is always the opposite sign and at most the same magnitude as the first omitted term. Due to Robbins, we can bound:
$$\sqrt{2\pi }n^{n+1/2}e^{-n} e^{\frac{1}{12n+1}} < n! < \sqrt{2\pi }n^{n+1/2} e^{−n} e^{1/12n}$$
More on Stirling Series in Base $2$
Let's develop the question of Stirling series when we have a base $2$ for example. The above approximation has to be read this way:
$$log_2(N!) \approx \log_2(\sqrt{2\pi N} N^N\ e^{-N})$$
Due to the fact that we have a non-natural log, it becomes
$$\log_2(N!) \approx \frac{1}{2}\log_2(2\pi N) + N\log_2(N) - N\log_2(e)$$
Hence one has to be very careful with the last term which is not $N$ anymore, but $N\log_2(e)$.
That being said one can proceed with the rest of Stirling series.
See the comments for numerical results.
Beauty Report
$$\color{red}{256\log_2(256) - 256\log_2(e) + \frac{1}{2}\log_2(2\pi\cdot 256) = 1683.9958175971615}$$
a very good accord with numerical evaluation (for example W. Mathematica) which gives $\log_2(256!) = 1683.9962872242145$.
Just a comment:
Of course, there are many calculators that can handle $\log_2 256!$ and much "worse" expressions directly. For example PARI/GP, if you type
log(256!)/log(2)
you will get a result like:
%1 = 1683.9962872242146194061476793149931595
(the number of digits can be configured with the default called realprecision).
If you want an exact integer logarithm, you can also use logint(256!, 2) which will give you 1683.
Typing 256! alone will give you the full 507-digit decimal expansion of this integer.
If PARI/GP is allowed to use memory (set parisizemax default), it will also immediately say that logint(654321!, 2) is 11697270.
As noted in comment, with reference to answer by Charles, if you want to work with floating-point operations (and not huge exact integers), you can use function lngamma which is equal to $\log\circ\Gamma$ for positive real arguments. Remember that compared to factorial, the Gamma function is shifted by one. So
$$\log_2 n! = \frac{\log n!}{\log 2} = \frac{\log \Gamma(n+1)}{\log 2}$$
and you can type lngamma(654321 + 1)/log(2) in PARI/GP and everything will be floating point operations. This will work for astronomical values of $n$, for example lngamma(3.14e1000) is fine ($\log\Gamma(3.14\cdot 10^{1000})$).