Try this link Casio Web Calculator

Maybe the difference between $$$log2(2^{59}-1)$$$ and $$$log2(2^{59})$$$ are so small that the answer is automatically adjusted to $$$59$$$

if you use the calculator on Windows 10 you can get 58.999999999999999997497323043895

Write a simple loop and count how often you need to multiply by 2.

Simply implement it (for integer answers):

int ilog(long long x){
    for(int i=0;i<=70;i++){
        if(1ll<<i >= x){
            return i;//if you want to ceil the answer, use this version
        }
        if(1ll<<i > x){
            return i-1;//if you want to floor the answer, use this version
        }
    }
}

Interesting, I've never seen that even though it seems more direct (I don't actually use C++ so it's not that surprising, I guess).

In Java for the other five Java users: Long.numberOfTrailingZeros(Long.highestOneBit(x)).

What's the difference between log2 and __lg?

Only since C++20 though?

Sure, floating point numbers store first several digits, depending on the number of bits provided for that (mantissa), and the other bunch of bits show the decimal point location (exponent).

For instance, for double it stores only 14-15 digits, and $$$2^{59}$$$ has more than that, the last digit isn't being stored.

P.S. Actually, it's more complex than what I said, because it doesn't store decimal digits, it stores in binary, that's why, for example, $$$2^{59}$$$ is printed correctly https://ideone.com/dPsCa6

Wikipedia has a pretty good description of the double format.