Any old scientific calculator can tell you that log(2) is approximately equal to .301030. The log of 2 taken to the power of one million (2^1000000) is log(2)*1,000,000, or 301030. It's not trivial to use this number to determine 2^1,000,000 down to the last digit, but it IS trivial to say how _many_ digits the number has. It is: 301,031 digits (because 10^0 is equal to 1, we need to add 1 to our result, then throw away the non-integer part).
For a simple verification of my assertion, let's take a look at 2^10, which is easy to compute: it's 1024. 10 times .301030 is 3.01030. Adding 1 to this is 4.01030. Taking the integer value we get 4, and that's how many digits 1024 has.
In fact, we can use this scheme for any power of 2. For example, we know that 2^16 = 65536. 16 times .301030 + 1 = 5.816, so we verify that the scheme works for any power of 2. The limitation to the approach is the accuracy of the value we use for log(2), which is a transcendental number so it has an infinite number of digits. But there are some online calculators we can use to get quite a few more digits. One reports that log10(2) = 0.30102999566398114, which should suffice for determining how many digits 2^N has up to N = 10^17. According to an earlier post of mine we already know that its least-significant digit is a 6 😀.
By the way, we know that log(2) MUST be a transcendental number because the relationship I described above has to hold for any arbitrarily-large power of N for 2^N. It doesn't matter if N = 10 or 10^10^10....
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