如何在Golang中计算256位整数的log16

Question

how to get the log with base 16 for a math/big Int variable.

Any help would be great since I am new to Go and came from Python and C environment

s := "c6d86e5a2cb4bc532361c2d4940f0b1a0138066e25d65c1c530d080b11f8ca24" // Hex value
i := new(big.Int)
i.SetString(s, 16) // hex value to Big Int
// how to get the log with base 16 for a math/big Int variable.

How it works in python

import math 
a = 0xc6d86e5a2cb4bc532361c2d4940f0b1a0138066e25d65c1c530d080b11f8ca24
a>> 89940344608680314083397671686667731393131665861770496634981932531495305005604L
math.log(a)/math.log(16.0)

answer turns out to be 63.908875905794794

Answer 1

An interesting property of logarithms is that change-of-base is actually pretty easy.

log_b (x) = log_a (x) / log_a (b)

So if you want to get log_16 (x), you could use the Log function and do change of base:

log_e (x) = log_16 (x) / log_16 (e)
log_16 (e) = approximately 0.36067 
=> log_16 (x) = 0.36067 * log_e (x)

So in Go, I think this would be:

li := Log(i) * 0.36067

Edit: When I wrote the answer above I did not realize that Log wouldn't work on a Big Int. Reading the Go Github, it looks like this is a requested feature in the language which has not yet been implemented due to lack of a satisfactorily quick solution. From what I read it looks like the best solution for the moment, if you must use a Big Int, is probably a Taylor Series implementation, which in my opinion would be non-trivial to write. The thread I linked to indicates that one or more such implementations may exist but are not necessarily computationally correct.