In case you don't want to return a BigInt, then the following might work for you too:

function log10(bigint) {
  if (bigint < 0) return NaN;
  const s = bigint.toString(10);

  return s.length + Math.log10("0." + s.substring(0, 15))
}

function log(bigint) {
  return log10(bigint) * Math.log(10);
}

function natlog(bigint) {
  if (bigint < 0) return NaN;

  const s = bigint.toString(16);
  const s15 = s.substring(0, 15);

  return Math.log(16) * (s.length - s15.length) + Math.log("0x" + s15);
}

const largeNumber = BigInt('9039845039485903949384755723427863486200719925474009384509283489374539477777093824750398247503894750384750238947502389475029384755555555555555555555555555555555555555554444444444444444444444444222222222222222222222255666666666666938475938475938475938408932475023847502384750923847502389475023987450238947509238475092384750923847502389457028394750293847509384570238497575938475938475938475938475555555555559843991');

console.log(natlog(largeNumber)); // 948.5641152531601
console.log(log10(largeNumber), log(largeNumber), log(-1))
// 411.95616098588766
// 948.5641152531603
// NaN

log10() will return a standard precision float for any BigInt or Int number you enter as an argument.

As @Mielipuoli quite rightly mentioned, the natural logarithm can be calculated as

function log(bigint) {
  return log10(bigint) / Math.log10(Math.E);
}

Or, even simpler, as shown in my snippet above, as log10(bigint) * Math.log(10).

@Nat already explained in a comment below, how this approach works, i.e. by calculating the integer and fractional parts of the logarithm separately and summing them up. With regards to the precision of the result: the Math.log10() works on a float number with its usual 13 to 14 decimal digits precision, and so, for a result, this is all you can expect too.

For this reason, I truncated the string representation of the BigInt number to 15 characters. Any further decimal places would have been ignored in the implicit type conversion to float anyway.

I also added the hex-string version here, suggested by @PeterCordes and further developed by @somebody as natlog(). It works - probably faster than my original solution - and produces the "same" result (only the very last shown digit deviates between the two results)!

The other answers have adequately addressed the question you give in the title, viz.: "how do I compute the logarithm of a BigInt?". However, you also mention that you are particularly interested in logarithms of factorials, for which a different algorithm avoids your range difficulties.

Applying log(ab) = log(a) + log(b), the following function computes the log of a factorial:

function logFactorial(n) {
  let total = 0;
  for (let current = 1; current <= n; ++current) {
    total += Math.log10(current);
  }

  return total;
}

console.log(logFactorial(170));

Inspired from MWO's answer, you could simply convert the BigInt into a string with the same base as the logarithm that you want to calculate and get the string length.

For example to calculate floor(log2(9007199254740991)) you can do BigInt("9007199254740991").toString(2).length - 1.

Note that toString only allows bases from 2 to 36.